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Partial derivatives and the gradient

Differentiating in one direction at a time; the gradient as the vector of slopes.

∂f/∂x freezes y and differentiates in x. The gradient ∇f = (∂f/∂x, ∂f/∂y) points uphill and is perpendicular to the level curves. Picture it: stand on the surface; the gradient is the direction water would flow uphill, and its length is the steepness. Think it: the gradient is the derivative — the unique linear map that best approximates f near the point.

Derivatives of a Function of Two Variables

When studying derivatives of functions of one variable, we found that one interpretation of the derivative is an instantaneous rate of change of \(y\) as a function of \(x.\) Leibniz notation for the derivative is \(dy\text{/}dx,\) which implies that \(y\) is the dependent variable and \(x\) is the independent variable. For a function \(z=f(x,y)\) of two variables, \(x\) and \(y\) are the independent variables and \(z\) is the dependent variable. This raises two questions right away: How do we adapt Leibniz notation for functions of two variables? Also, what is an interpretation of the derivative? The answer lies in partial derivatives.

This definition shows two differences already. First, the notation changes, in the sense that we still use a version of Leibniz notation, but the \(d\) in the original notation is replaced with the symbol \(∂.\) (This rounded \(\text{“d”}\) is usually called “partial,” so \(∂f\text{/}∂x\) is spoken as the “partial of \(f\) with respect to \(x\text{.”})\) This is the first hint that we are dealing with partial derivatives. Second, we now have two different derivatives we can take, since there are two different independent variables. Depending on which variable we choose, we can come up with different partial derivatives altogether, and often do.

The idea to keep in mind when calculating partial derivatives is to treat all independent variables, other than the variable with respect to which we are differentiating, as constants. Then proceed to differentiate as with a function of a single variable. To see why this is true, first fix \(y\) and define \(g(x)=f(x,y)\) as a function of \(x.\) Then

\[{g}^{'}(x)=\underset{h\to 0}{\text{lim}}\frac{g(x+h)-g(x)}{h}=\underset{h\to 0}{\text{lim}}\frac{f(x+h,y)-f(x,y)}{h}=\frac{∂f}{∂x}.\]

The same is true for calculating the partial derivative of \(f\) with respect to \(y.\) This time, fix \(x\) and define \(h(y)=f(x,y)\) as a function of \(y.\) Then

\[{h}^{'}(x)=\underset{k\to 0}{\text{lim}}\frac{h(x+k)-h(x)}{k}=\underset{k\to 0}{\text{lim}}\frac{f(x,y+k)-f(x,y)}{k}=\frac{∂f}{∂y}.\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Functions of More Than Two Variables

Suppose we have a function of three variables, such as \(w=f(x,y,z).\) We can calculate partial derivatives of \(w\) with respect to any of the independent variables, simply as extensions of the definitions for partial derivatives of functions of two variables.

We can calculate a partial derivative of a function of three variables using the same idea we used for a function of two variables. For example, if we have a function \(f\) of \(x,y,\text{and}\ z,\) and we wish to calculate \(∂f\text{/}∂x,\) then we treat the other two independent variables as if they are constants, then differentiate with respect to \(x.\)

Condensed — the full section is in OpenStax Calculus Volume 3.

Higher-Order Partial Derivatives

Consider the function

\[f(x,y)=2{x}^{3}-4x{y}^{2}+5{y}^{3}-6xy+5x-4y+12.\]

Its partial derivatives are

\[\frac{∂f}{∂x}=6{x}^{2}-4{y}^{2}-6y+5\ \text{and}\ \frac{∂f}{∂y}=-8xy+15{y}^{2}-6x-4.\]

Each of these partial derivatives is a function of two variables, so we can calculate partial derivatives of these functions. Just as with derivatives of single-variable functions, we can call these second-order derivatives, third-order derivatives, and so on. In general, they are referred to as higher-order partial derivatives. There are four second-order partial derivatives for any function (provided they all exist):

\[\frac{{∂}^{2}f}{∂{x}^{2}}=\frac{∂}{∂x}[\frac{∂f}{∂x}],\ \frac{{∂}^{2}f}{∂x∂y}=\frac{∂}{∂x}[\frac{∂f}{∂y}],\ \frac{{∂}^{2}f}{∂y∂x}=\frac{∂}{∂y}[\frac{∂f}{∂x}],\ \frac{{∂}^{2}f}{∂{y}^{2}}=\frac{∂}{∂y}[\frac{∂f}{∂y}].\]

An alternative notation for each is \({f}_{xx},{f}_{yx},{f}_{xy},\) and \({f}_{yy},\) respectively. Higher-order partial derivatives calculated with respect to different variables, such as \({f}_{xy}\) and \({f}_{yx},\) are commonly called mixed partial derivatives.

Example

Try it.

Calculate all four second partial derivatives for the function

\[f(x,y)=x{e}^{-3y}+\text{sin}(2x-5y).\]
Solution

To calculate \({∂}^{2}f\text{/}d{x}^{2}\) and \({∂}^{2}f\text{/}∂y∂x,\) we first calculate \(∂f\text{/}∂x\text{:}\)

\[\frac{∂f}{∂x}={e}^{-3y}+2\ \text{cos}(2x-5y).\]

To calculate \({∂}^{2}f\text{/}d{x}^{2},\) differentiate \(∂f\text{/}∂x\) with respect to \(x\text{:}\)

\[\begin{array}{ll}\frac{{∂}^{2}f}{∂{x}^{2}} & =\frac{∂}{∂x}[\frac{∂f}{∂x}] \\ & =\frac{∂}{∂x}[{e}^{-3y}+2\ \text{cos}(2x-5y)] \\ & =-4\ \text{sin}(2x-5y).\end{array}\]

To calculate \({∂}^{2}f\text{/}∂y∂x,\) differentiate \(∂f\text{/}∂x\) with respect to \(y\text{:}\)

\[\begin{array}{ll}\frac{{∂}^{2}f}{∂y∂x} & =\frac{∂}{∂y}[\frac{∂f}{∂x}] \\ & =\frac{∂}{∂y}[{e}^{-3y}+2\ \text{cos}(2x-5y)] \\ & =-3{e}^{-3y}+10\ \text{sin}(2x-5y).\end{array}\]

To calculate \({∂}^{2}f\text{/}∂x∂y\) and \({∂}^{2}f\text{/}d{y}^{2},\) first calculate \(∂f\text{/}∂y\text{:}\)

\[\frac{∂f}{∂y}=-3x{e}^{-3y}-5\ \text{cos}(2x-5y).\]

To calculate \({∂}^{2}f\text{/}∂x∂y,\) differentiate \(∂f\text{/}∂y\) with respect to \(x\text{:}\)

\[\begin{array}{ll}\frac{{∂}^{2}f}{∂x∂y} & =\frac{∂}{∂x}[\frac{∂f}{∂y}] \\ & =\frac{∂}{∂x}[-3x{e}^{-3y}-5\ \text{cos}(2x-5y)] \\ & =-3{e}^{-3y}+10\ \text{sin}(2x-5y).\end{array}\]

To calculate \({∂}^{2}f\text{/}∂{y}^{2},\) differentiate \(∂f\text{/}∂y\) with respect to \(y\text{:}\)

\[\begin{array}{ll}\frac{{∂}^{2}f}{∂{y}^{2}} & =\frac{∂}{∂y}[\frac{∂f}{∂y}] \\ & =\frac{∂}{∂y}[-3x{e}^{-3y}-5\ \text{cos}(2x-5y)] \\ & =9x{e}^{-3y}-25\ \text{sin}(2x-5y).\end{array}\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Partial Differential Equations

In Introduction to Differential Equations, we studied differential equations in which the unknown function had one independent variable. A partial differential equation is an equation that involves an unknown function of more than one independent variable and one or more of its partial derivatives. Examples of partial differential equations are

\[{u}_{t}={c}^{2}({u}_{xx}+{u}_{yy})\]

(heat equation in two dimensions)

\[{u}_{tt}={c}^{2}({u}_{xx}+{u}_{yy})\]

(wave equation in two dimensions)

\[{u}_{xx}+{u}_{yy}=0\]

(Laplace’s equation in two dimensions)

In the first two equations, the unknown function \(u\) has three independent variables—\(t,x,\ \text{and}\ y\)—and \(c\) is an arbitrary constant. The independent variables \(x\ \text{and}\ y\) are considered to be spatial variables, and the variable \(t\) represents time. In Laplace’s equation, the unknown function \(u\) has two independent variables \(x\ \text{and}\ y.\)

Example

Try it.

Verify that

\[u(x,y,t)=5\ \text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t)\]

is a solution to the wave equation

\[{u}_{tt}=4({u}_{xx}+{u}_{yy}).\]
Solution

First, we calculate \({u}_{tt},{u}_{xx},\) and \({u}_{yy}\text{:}\)

\[\begin{array}{lll}{u}_{tt} & = & \frac{∂}{∂t}[\frac{∂u}{∂t}] \\ & = & \frac{∂}{∂t}[5\ \text{sin}(3\pi x)\text{sin}(4\pi y)(-10\pi \ \text{sin}(10\pi t))] \\ & = & \frac{∂}{∂t}[-50\pi \ \text{sin}(3\pi x)\text{sin}(4\pi y)\text{sin}(10\pi t)] \\ & = & -500{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t) \\ {u}_{xx} & = & \frac{∂}{∂x}[\frac{∂u}{∂x}] \\ & = & \frac{∂}{∂x}[15\pi \ \text{cos}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t)] \\ & = & -45{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t) \\ {u}_{yy} & = & \frac{∂}{∂y}[\frac{∂u}{∂y}] \\ & = & \frac{∂}{∂y}[5\ \text{sin}(3\pi x)(4\pi \ \text{cos}(4\pi y))\text{cos}(10\pi t)] \\ & = & \frac{∂}{∂y}[20\pi \ \text{sin}(3\pi x)\text{cos}(4\pi y)\text{cos}(10\pi t)] \\ & = & -80{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t).\end{array}\]

Next, we substitute each of these into the right-hand side of and simplify:

\[\begin{array}{ll}4({u}_{xx}+{u}_{yy}) & =4(-45{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t)+-80{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t)) \\ & =4(-125{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t)) \\ & =-500{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t) \\ & ={u}_{tt}.\end{array}\]

This verifies the solution.

Condensed — the full section is in OpenStax Calculus Volume 3.

Key Concepts

  • A partial derivative is a derivative involving a function of more than one independent variable.
  • To calculate a partial derivative with respect to a given variable, treat all the other variables as constants and use the usual differentiation rules.
  • Higher-order partial derivatives can be calculated in the same way as higher-order derivatives.

Key Equations

Partial derivative of \(f\) with respect to \(x\)\(\frac{∂f}{∂x}=\underset{h\to 0}{\text{lim}}\frac{f(x+h,y)-f(x,y)}{h}\)
Partial derivative of \(f\) with respect to \(y\)\(\frac{∂f}{∂y}=\underset{k\to 0}{\text{lim}}\frac{f(x,y+k)-f(x,y)}{k}\)

Partial Derivatives

For the following exercises, calculate the partial derivative using the limit definitions only.

For the following exercises, calculate the sign of the partial derivative using the graph of the surface.

For the following exercises, calculate the partial derivatives.

Find the indicated higher-order partial derivatives.

പ്രവര്‍ത്തിപ്പിച്ച ഉദാഹരണം: gradient of x^2*y - 3x + y^3

Multivariable-Calculus · Partial-Derivatives

Gradient of x^2y - 3x + y^3

x^{2} y - 3 x + y^{3}

പടിയിലൂടെ

  1. f(x, y) = x^{2} y - 3 x + y^{3}

    The gradient is the vector of partial derivatives — differentiate with respect to each variable, holding the others constant.

  2. \frac{\partial f}{\partial x} = 2 x y - 3

    Treat every variable except x as a constant.

  3. \frac{\partial f}{\partial y} = x^{2} + 3 y^{2}

    Treat every variable except y as a constant.

  4. \nabla f = \left[\begin{matrix}2 x y - 3\\x^{2} + 3 y^{2}\end{matrix}\right]

    Assemble the gradient vector. It points in the direction of steepest ascent.

ഉത്തരം വെളിപ്പെടുത്തുക
\nabla f = \left[\begin{matrix}2 x y - 3\\x^{2} + 3 y^{2}\end{matrix}\right]

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Use the definition of the partial derivative as a limit to calculate \(∂f\text{/}∂x\) and \(∂f\text{/}∂y\) for the function

    \[f(x,y)={x}^{2}-3xy+2{y}^{2}-4x+5y-12.\]
    ഉത്തരം വെളിപ്പെടുത്തുക

    First, calculate \(f(x+h,y).\)

    \[\begin{array}{ll}f(x+h,y) & ={(x+h)}^{2}-3(x+h)y+2{y}^{2}-4(x+h)+5y-12 \\ & ={x}^{2}+2xh+{h}^{2}-3xy-3hy+2{y}^{2}-4x-4h+5y-12.\end{array}\]

    Next, substitute this into and simplify:

    \[\begin{array}{ll}\frac{∂f}{∂x} & =\underset{h\to 0}{\text{lim}}\frac{f(x+h,y)-f(x,y)}{h} \\ & =\underset{h\to 0}{\text{lim}}\frac{({x}^{2}+2xh+{h}^{2}-3xy-3hy+2{y}^{2}-4x-4h+5y-12)-({x}^{2}-3xy+2{y}^{2}-4x+5y-12)}{h} \\ & =\underset{h\to 0}{\text{lim}}\frac{{x}^{2}+2xh+{h}^{2}-3xy-3hy+2{y}^{2}-4x-4h+5y-12-{x}^{2}+3xy-2{y}^{2}+4x-5y+12}{h} \\ & =\underset{h\to 0}{\text{lim}}\frac{2xh+{h}^{2}-3hy-4h}{h} \\ & =\underset{h\to 0}{\text{lim}}\frac{h(2x+h-3y-4)}{h} \\ & =\underset{h\to 0}{\text{lim}}(2x+h-3y-4) \\ & =2x-3y-4.\end{array}\]

    To calculate \(\frac{∂f}{∂y},\) first calculate \(f(x,y+k)\text{:}\)

    \[\begin{array}{ll}f(x,y+k) & ={x}^{2}-3x(y+k)+2{(y+k)}^{2}-4x+5(y+k)-12 \\ & ={x}^{2}-3xy-3xk+2{y}^{2}+4yk+2{k}^{2}-4x+5y+5k-12.\end{array}\]

    Next, substitute this into and simplify:

    \[\begin{array}{ll}\frac{∂f}{∂y} & =\underset{k\to 0}{\text{lim}}\frac{f(x,y+k)-f(x,y)}{k} \\ & =\underset{k\to 0}{\text{lim}}\frac{({x}^{2}-3xy-3xk+2{y}^{2}+4yk+2{k}^{2}-4x+5y+5k-12)-({x}^{2}-3xy+2{y}^{2}-4x+5y-12)}{k} \\ & =\underset{k\to 0}{\text{lim}}\frac{{x}^{2}-3xy-3xk+2{y}^{2}+4yk+2{k}^{2}-4x+5y+5k-12-{x}^{2}+3xy-2{y}^{2}+4x-5y+12}{k} \\ & =\underset{k\to 0}{\text{lim}}\frac{-3xk+4yk+2{k}^{2}+5k}{k} \\ & =\underset{k\to 0}{\text{lim}}\frac{h(-3x+4y+2k+5)}{k} \\ & =\underset{k\to 0}{\text{lim}}(-3x+4y+2k+5) \\ & =-3x+4y+5.\end{array}\]

  2. Use the definition of the partial derivative as a limit to calculate \(∂f\text{/}∂x\) and \(∂f\text{/}∂y\) for the function

    \[f(x,y)=4{x}^{2}+2xy-{y}^{2}+3x-2y+5.\]
    ഉത്തരം വെളിപ്പെടുത്തുക

    \(\frac{∂f}{∂x}=8x+2y+3,\ \frac{∂f}{∂y}=2x-2y-2\)

  3. Calculate both partial derivatives for the following functions by holding the opposite variable constant then differentiating:

    1. \(f(x,y)={x}^{2}-3xy+2{y}^{2}-4x+5y-12\)
    2. \(g(x,y)=\text{sin}({x}^{2}y-2x+4)\)
    ഉത്തരം വെളിപ്പെടുത്തുക
    1. To calculate \(∂f\text{/}∂x,\) treat the variable \(y\) as a constant. Then differentiate \(f(x,y)\) with respect to \(x\) using the sum, difference, and power rules:
      \[\begin{array}{ll}\frac{∂f}{∂x} & =\frac{∂}{∂x}[{x}^{2}-3xy+2{y}^{2}-4x+5y-12] \\ & =\frac{∂}{∂x}[{x}^{2}]-\frac{∂}{∂x}[3xy]+\frac{∂}{∂x}[2{y}^{2}]-\frac{∂}{∂x}[4x]+\frac{∂}{∂x}[5y]-\frac{∂}{∂x}[12] \\ & =2x-3y+0-4+0-0 \\ & =2x-3y-4.\end{array}\]
      The derivatives of the third, fifth, and sixth terms are all zero because they do not contain the variable \(x,\) so they are treated as constant terms. The derivative of the second term is equal to the coefficient of \(x,\) which is \(-3y.\) Calculating \(∂f\text{/}∂y\text{:}\)
      \[\begin{array}{ll}\frac{∂f}{∂y} & =\frac{∂}{∂y}[{x}^{2}-3xy+2{y}^{2}-4x+5y-12] \\ & =\frac{∂}{∂y}[{x}^{2}]-\frac{∂}{∂y}[3xy]+\frac{∂}{∂y}[2{y}^{2}]-\frac{∂}{∂y}[4x]+\frac{∂}{∂y}[5y]-\frac{∂}{∂y}[12] \\ & =-3x+4y-0+5-0 \\ & =-3x+4y+5.\end{array}\]
      These are the same answers obtained in .
    2. To calculate \(∂g\text{/}∂x,\) treat the variable y as a constant. Then differentiate \(g(x,y)\) with respect to x using the chain rule and power rule:
      \[\begin{array}{ll}\frac{∂g}{∂x} & =\frac{∂}{∂x}[\text{sin}({x}^{2}y-2x+4)] \\ & =\text{cos}({x}^{2}y-2x+4)\frac{∂}{∂x}[{x}^{2}y-2x+4] \\ & =(2xy-2)\text{cos}({x}^{2}y-2x+4).\end{array}\]
      To calculate \(∂g\text{/}∂y,\) treat the variable \(x\) as a constant. Then differentiate \(g(x,y)\) with respect to \(y\) using the chain rule and power rule:
      \[\begin{array}{ll}\frac{∂g}{∂y} & =\frac{∂}{∂y}[\text{sin}({x}^{2}y-2x+4)] \\ & =\text{cos}({x}^{2}y-2x+4)\frac{∂}{∂y}[{x}^{2}y-2x+4] \\ & ={x}^{2}\text{cos}({x}^{2}y-2x+4).\end{array}\]

  4. Calculate \(∂f\text{/}∂x\) and \(∂f\text{/}∂y\) for the function \(f(x,y)=\text{tan}({x}^{3}-3{x}^{2}{y}^{2}+2{y}^{4})\) by holding the opposite variable constant, then differentiating.

    ഉത്തരം വെളിപ്പെടുത്തുക

    \(\begin{array}{l}\frac{∂f}{∂x}=(3{x}^{2}-6x{y}^{2}){\text{sec}}^{2}({x}^{3}-3{x}^{2}{y}^{2}+2{y}^{4}) \\ \frac{∂f}{∂y}=(-6{x}^{2}y+8{y}^{3}){\text{sec}}^{2}({x}^{3}-3{x}^{2}{y}^{2}+2{y}^{4})\end{array}\)

  5. Use a contour map to estimate \(∂g\text{/}∂x\) at the point \((\sqrt{5},0)\) for the function \(g(x,y)=\sqrt{9-{x}^{2}-{y}^{2}}.\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    The following graph represents a contour map for the function \(g(x,y)=\sqrt{9-{x}^{2}-{y}^{2}}.\)

    The inner circle on the contour map corresponds to \(c=2\) and the next circle out corresponds to \(c=1.\) The first circle is given by the equation \(2=\sqrt{9-{x}^{2}-{y}^{2}};\) the second circle is given by the equation \(1=\sqrt{9-{x}^{2}-{y}^{2}}.\) The first equation simplifies to \({x}^{2}+{y}^{2}=5\) and the second equation simplifies to \({x}^{2}+{y}^{2}=8.\) The \(x\text{-intercept}\) of the first circle is \((\sqrt{5},0)\) and the \(x\text{-intercept}\) of the second circle is \((2\sqrt{2},0).\) We can estimate the value of \(∂g\text{/}∂x\) evaluated at the point \((\sqrt{5},0)\) using the slope formula:

    \[{\frac{∂g}{∂x}|}_{(x,y)=(\sqrt{5},0)}\approx \frac{g(\sqrt{5},0)-g(2\sqrt{2},0)}{\sqrt{5}-2\sqrt{2}}=\frac{2-1}{\sqrt{5}-2\sqrt{2}}=\frac{1}{\sqrt{5}-2\sqrt{2}}\approx -1.688.\]

    To calculate the exact value of \(∂g\text{/}∂x\) evaluated at the point \((\sqrt{5},0),\) we start by finding \(∂g\text{/}∂x\) using the chain rule. First, we rewrite the function as \(g(x,y)=\sqrt{9-{x}^{2}-{y}^{2}}={(9-{x}^{2}-{y}^{2})}^{1\text{/}2}\) and then differentiate with respect to \(x\) while holding \(y\) constant:

    \[\frac{∂g}{∂x}=\frac{1}{2}{(9-{x}^{2}-{y}^{2})}^{-1\text{/}2}(-2x)=-\frac{x}{\sqrt{9-{x}^{2}-{y}^{2}}}.\]

    Next, we evaluate this expression using \(x=\sqrt{5}\) and \(y=0\text{:}\)

    \[{\frac{∂g}{∂x}|}_{(x,y)=(\sqrt{5},0)}=-\frac{\sqrt{5}}{\sqrt{9-{(\sqrt{5})}^{2}-{(0)}^{2}}}=-\frac{\sqrt{5}}{\sqrt{4}}=-\frac{\sqrt{5}}{2}\approx -1.118.\]

    The estimate for the partial derivative corresponds to the slope of the secant line passing through the points \((\sqrt{5},0,g(\sqrt{5},0))\) and \((2\sqrt{2},0,g(2\sqrt{2},0)).\) It represents an approximation to the slope of the tangent line to the surface through the point \((\sqrt{5},0,g(\sqrt{5},0)),\) which is parallel to the \(x\text{-axis}.\)

  6. Use a contour map to estimate \(∂f\text{/}∂y\) at point \((0,\sqrt{2})\) for the function

    \[f(x,y)={x}^{2}-{y}^{2}.\]

    Compare this with the exact answer.

    ഉത്തരം വെളിപ്പെടുത്തുക

    Using the curves corresponding to \(c=-2\ \text{and}\ c=-3,\) we obtain

    \({\frac{∂f}{∂y}|}_{(x,y)=(0,\sqrt{2})}\approx \frac{f(0,\sqrt{3})-f(0,\sqrt{2})}{\sqrt{3}-\sqrt{2}}=\frac{-3-(-2)}{\sqrt{3}-\sqrt{2}}\cdot \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\text{-}\sqrt{3}-\sqrt{2}\approx -3.146.\)

    The exact answer is

    \({\frac{∂f}{∂y}|}_{(x,y)=(0,\sqrt{2})}=({-2y|}_{(x,y)=(0,\sqrt{2})}=-2\sqrt{2}\approx -2.828.\)

  7. Use the limit definition of partial derivatives to calculate \(∂f\text{/}∂x\) for the function

    \[f(x,y,z)={x}^{2}-3xy+2{y}^{2}-4xz+5y{z}^{2}-12x+4y-3z.\]

    Then, find \(∂f\text{/}∂y\) and \(∂f\text{/}∂z\) by setting the other two variables constant and differentiating accordingly.

    ഉത്തരം വെളിപ്പെടുത്തുക

    We first calculate \(∂f\text{/}∂x\) using , then we calculate the other two partial derivatives by holding the remaining variables constant. To use the equation to find \(∂f\text{/}∂x,\) we first need to calculate \(f(x+h,y,z)\text{:}\)

    \[\begin{array}{ll}f(x+h,y,z) & ={(x+h)}^{2}-3(x+h)y+2{y}^{2}-4(x+h)z+5y{z}^{2}-12(x+h)+4y-3z \\ & ={x}^{2}+2xh+{h}^{2}-3xy-3xh+2{y}^{2}-4xz-4hz+5y{z}^{2}-12x-12h+4y-3z\end{array}\]

    and recall that \(f(x,y,z)={x}^{2}-3xy+2{y}^{2}-4zx+5y{z}^{2}-12x+4y-3z.\) Next, we substitute these two expressions into the equation:

    \[\begin{array}{ll}\frac{∂f}{∂x} & =\underset{h\to 0}{\text{lim}}[\frac{{x}^{2}+2xh+{h}^{2}-3xy-3hy+2{y}^{2}-4xz-4hz+5y{z}^{2}-12x-12h+4y-3z}{h} \\ & \ -\frac{{x}^{2}-3xy+2{y}^{2}-4xz+5y{z}^{2}-12x+4y-3z}{h}] \\ & =\underset{h\to 0}{\text{lim}}[\frac{2xh+{h}^{2}-3hy-4hz-12h}{h}] \\ & =\underset{h\to 0}{\text{lim}}[\frac{h(2x+h-3y-4z-12)}{h}] \\ & =\underset{h\to 0}{\text{lim}}(2x+h-3y-4z-12) \\ & =2x-3y-4z-12.\end{array}\]

    Then we find \(∂f\text{/}∂y\) by holding \(x\ \text{and}\ z\) constant. Therefore, any term that does not include the variable \(y\) is constant, and its derivative is zero. We can apply the sum, difference, and power rules for functions of one variable:

    \[\begin{array}{l}\frac{∂}{∂y}[{x}^{2}-3xy+2{y}^{2}-4xz+5y{z}^{2}-12x+4y-3z] \\ =\frac{∂}{∂y}[{x}^{2}]-\frac{∂}{∂y}[3xy]+\frac{∂}{∂y}[2{y}^{2}]-\frac{∂}{∂y}[4xz]+\frac{∂}{∂y}[5y{z}^{2}]-\frac{∂}{∂y}[12x]+\frac{∂}{∂y}[4y]-\frac{∂}{∂y}[3z] \\ =0-3x+4y-0+5{z}^{2}-0+4-0 \\ =-3x+4y+5{z}^{2}+4.\end{array}\]

    To calculate \(∂f\text{/}∂z,\) we hold x and y constant and apply the sum, difference, and power rules for functions of one variable:

    \[\begin{array}{l}\frac{∂}{∂z}[{x}^{2}-3xy+2{y}^{2}-4xz+5y{z}^{2}-12x+4y-3z] \\ =\frac{∂}{∂z}[{x}^{2}]-\frac{∂}{∂z}[3xy]+\frac{∂}{∂z}[2{y}^{2}]-\frac{∂}{∂z}[4xz]+\frac{∂}{∂z}[5y{z}^{2}]-\frac{∂}{∂z}[12x]+\frac{∂}{∂z}[4y]-\frac{∂}{∂z}[3z] \\ =0-0+0-4x+10yz-0+0-3 \\ =-4x+10yz-3.\end{array}\]

  8. Use the limit definition of partial derivatives to calculate \(∂f\text{/}∂x\) for the function

    \[f(x,y,z)=2{x}^{2}-4{x}^{2}y+2{y}^{2}+5x{z}^{2}-6x+3z-8.\]

    Then find \(∂f\text{/}∂y\) and \(∂f\text{/}∂z\) by setting the other two variables constant and differentiating accordingly.

    ഉത്തരം വെളിപ്പെടുത്തുക

    \(\frac{∂f}{∂x}=4x-8xy+5{z}^{2}-6,\ \frac{∂f}{∂y}=-4{x}^{2}+4y,\ \frac{∂f}{∂z}=10xz+3\)

  9. Calculate the three partial derivatives of the following functions.

    1. \(f(x,y,z)=\frac{{x}^{2}y-4xz+{y}^{2}}{x-3yz}\)
    2. \(g(x,y,z)=\text{sin}({x}^{2}y-z)+\text{cos}({x}^{2}-yz)\)
    ഉത്തരം വെളിപ്പെടുത്തുക

    In each case, treat all variables as constants except the one whose partial derivative you are calculating.

    1. \(\begin{array}{ll}\frac{∂f}{∂x} & =\frac{∂}{∂x}[\frac{{x}^{2}y-4xz+{y}^{2}}{x-3yz}] \\ & =\frac{\frac{∂}{∂x}({x}^{2}y-4xz+{y}^{2})(x-3yz)-({x}^{2}y-4xz+{y}^{2})\frac{∂}{∂x}(x-3yz)}{{(x-3yz)}^{2}} \\ & =\frac{(2xy-4z)(x-3yz)-({x}^{2}y-4xz+{y}^{2})(1)}{{(x-3yz)}^{2}} \\ & =\frac{2{x}^{2}y-6x{y}^{2}z-4xz+12y{z}^{2}-{x}^{2}y+4xz-{y}^{2}}{{(x-3yz)}^{2}} \\ & =\frac{{x}^{2}y-6x{y}^{2}z-4xz+12y{z}^{2}+4xz-{y}^{2}}{{(x-3yz)}^{2}}\end{array}\)
      \(\begin{array}{ll}\frac{∂f}{∂y} & =\frac{∂}{∂y}[\frac{{x}^{2}y-4xz+{y}^{2}}{x-3yz}] \\ & =\frac{\frac{∂}{∂y}({x}^{2}y-4xz+{y}^{2})(x-3yz)-({x}^{2}y-4xz+{y}^{2})\frac{∂}{∂y}(x-3yz)}{{(x-3yz)}^{2}} \\ & =\frac{({x}^{2}+2y)(x-3yz)-({x}^{2}y-4xz+{y}^{2})(-3z)}{{(x-3yz)}^{2}} \\ & =\frac{{x}^{3}-3{x}^{2}yz+2xy-6{y}^{2}z+3{x}^{2}yz-12x{z}^{2}+3{y}^{2}z}{{(x-3yz)}^{2}} \\ & =\frac{{x}^{3}+2xy-3{y}^{2}z-12x{z}^{2}}{{(x-3yz)}^{2}}\end{array}\)
      \(\begin{array}{ll}\frac{∂f}{∂z} & =\frac{∂}{∂z}[\frac{{x}^{2}y-4xz+{y}^{2}}{x-3yz}] \\ & =\frac{\frac{∂}{∂z}({x}^{2}y-4xz+{y}^{2})(x-3yz)-({x}^{2}y-4xz+{y}^{2})\frac{∂}{∂z}(x-3yz)}{{(x-3yz)}^{2}} \\ & =\frac{(-4x)(x-3yz)-({x}^{2}y-4xz+{y}^{2})(-3y)}{{(x-3yz)}^{2}} \\ & =\frac{-4{x}^{2}+12xyz+3{x}^{2}{y}^{2}-12xyz+3{y}^{3}}{{(x-3yz)}^{2}} \\ & =\frac{-4{x}^{2}+3{x}^{2}{y}^{2}+3{y}^{3}}{{(x-3yz)}^{2}}\end{array}\)
    2. \(\begin{array}{ll}\frac{∂f}{∂x} & =\frac{∂}{∂x}[\text{sin}({x}^{2}y-z)+\text{cos}({x}^{2}-yz)] \\ & =(\text{cos}({x}^{2}y-z))\frac{∂}{∂x}({x}^{2}y-z)-(\text{sin}({x}^{2}-yz))\frac{∂}{∂x}({x}^{2}-yz) \\ & =2xy\ \text{cos}({x}^{2}y-z)-2x\ \text{sin}({x}^{2}-yz) \\ \frac{∂f}{∂y} & =\frac{∂}{∂y}[\text{sin}({x}^{2}y-z)+\text{cos}({x}^{2}-yz)] \\ & =(\text{cos}({x}^{2}y-z))\frac{∂}{∂y}({x}^{2}y-z)-(\text{sin}({x}^{2}-yz))\frac{∂}{∂y}({x}^{2}-yz) \\ & ={x}^{2}\text{cos}({x}^{2}y-z)+z\ \text{sin}({x}^{2}-yz) \\ \frac{∂f}{∂z} & =\frac{∂}{∂z}[\text{sin}({x}^{2}y-z)+\text{cos}({x}^{2}-yz)] \\ & =(\text{cos}({x}^{2}y-z))\frac{∂}{∂z}({x}^{2}y-z)-(\text{sin}({x}^{2}-yz))\frac{∂}{∂z}({x}^{2}-yz) \\ & =\text{-}\text{cos}({x}^{2}y-z)+y\ \text{sin}({x}^{2}-yz)\end{array}\)

  10. Calculate \(∂f\text{/}∂x,\) \(∂f\text{/}∂y,\) and \(∂f\text{/}∂z\) for the function \(f(x,y,z)=\text{sec}({x}^{2}y)-\text{tan}({x}^{3}y{z}^{2}).\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    \(\begin{array}{l}\frac{∂f}{∂x}=2xy\ \text{sec}({x}^{2}y)\text{tan}({x}^{2}y)-3{x}^{2}y{z}^{2}{\text{sec}}^{2}({x}^{3}y{z}^{2}) \\ \frac{∂f}{∂y}={x}^{2}\text{sec}({x}^{2}y)\text{tan}({x}^{2}y)-{x}^{3}{z}^{2}{\text{sec}}^{2}({x}^{3}y{z}^{2}) \\ \frac{∂f}{∂z}=-2{x}^{3}yz\ {\text{sec}}^{2}({x}^{3}y{z}^{2})\end{array}\)

  11. Calculate all four second partial derivatives for the function

    \[f(x,y)=x{e}^{-3y}+\text{sin}(2x-5y).\]
    ഉത്തരം വെളിപ്പെടുത്തുക

    To calculate \({∂}^{2}f\text{/}d{x}^{2}\) and \({∂}^{2}f\text{/}∂y∂x,\) we first calculate \(∂f\text{/}∂x\text{:}\)

    \[\frac{∂f}{∂x}={e}^{-3y}+2\ \text{cos}(2x-5y).\]

    To calculate \({∂}^{2}f\text{/}d{x}^{2},\) differentiate \(∂f\text{/}∂x\) with respect to \(x\text{:}\)

    \[\begin{array}{ll}\frac{{∂}^{2}f}{∂{x}^{2}} & =\frac{∂}{∂x}[\frac{∂f}{∂x}] \\ & =\frac{∂}{∂x}[{e}^{-3y}+2\ \text{cos}(2x-5y)] \\ & =-4\ \text{sin}(2x-5y).\end{array}\]

    To calculate \({∂}^{2}f\text{/}∂y∂x,\) differentiate \(∂f\text{/}∂x\) with respect to \(y\text{:}\)

    \[\begin{array}{ll}\frac{{∂}^{2}f}{∂y∂x} & =\frac{∂}{∂y}[\frac{∂f}{∂x}] \\ & =\frac{∂}{∂y}[{e}^{-3y}+2\ \text{cos}(2x-5y)] \\ & =-3{e}^{-3y}+10\ \text{sin}(2x-5y).\end{array}\]

    To calculate \({∂}^{2}f\text{/}∂x∂y\) and \({∂}^{2}f\text{/}d{y}^{2},\) first calculate \(∂f\text{/}∂y\text{:}\)

    \[\frac{∂f}{∂y}=-3x{e}^{-3y}-5\ \text{cos}(2x-5y).\]

    To calculate \({∂}^{2}f\text{/}∂x∂y,\) differentiate \(∂f\text{/}∂y\) with respect to \(x\text{:}\)

    \[\begin{array}{ll}\frac{{∂}^{2}f}{∂x∂y} & =\frac{∂}{∂x}[\frac{∂f}{∂y}] \\ & =\frac{∂}{∂x}[-3x{e}^{-3y}-5\ \text{cos}(2x-5y)] \\ & =-3{e}^{-3y}+10\ \text{sin}(2x-5y).\end{array}\]

    To calculate \({∂}^{2}f\text{/}∂{y}^{2},\) differentiate \(∂f\text{/}∂y\) with respect to \(y\text{:}\)

    \[\begin{array}{ll}\frac{{∂}^{2}f}{∂{y}^{2}} & =\frac{∂}{∂y}[\frac{∂f}{∂y}] \\ & =\frac{∂}{∂y}[-3x{e}^{-3y}-5\ \text{cos}(2x-5y)] \\ & =9x{e}^{-3y}-25\ \text{sin}(2x-5y).\end{array}\]

  12. Calculate all four second partial derivatives for the function

    \[f(x,y)=\text{sin}(3x-2y)+\text{cos}(x+4y).\]
    ഉത്തരം വെളിപ്പെടുത്തുക

    \(\begin{array}{l}\frac{{∂}^{2}f}{∂{x}^{2}}=-9\ \text{sin}(3x-2y)-\text{cos}(x+4y) \\ \frac{{∂}^{2}f}{∂x∂y}=6\ \text{sin}(3x-2y)-4\ \text{cos}(x+4y) \\ \frac{{∂}^{2}f}{∂y∂x}=6\ \text{sin}(3x-2y)-4\ \text{cos}(x+4y) \\ \frac{{∂}^{2}f}{∂{y}^{2}}=-4\ \text{sin}(3x-2y)-16\ \text{cos}(x+4y)\end{array}\)

  13. Verify that

    \[u(x,y,t)=5\ \text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t)\]

    is a solution to the wave equation

    \[{u}_{tt}=4({u}_{xx}+{u}_{yy}).\]
    ഉത്തരം വെളിപ്പെടുത്തുക

    First, we calculate \({u}_{tt},{u}_{xx},\) and \({u}_{yy}\text{:}\)

    \[\begin{array}{lll}{u}_{tt} & = & \frac{∂}{∂t}[\frac{∂u}{∂t}] \\ & = & \frac{∂}{∂t}[5\ \text{sin}(3\pi x)\text{sin}(4\pi y)(-10\pi \ \text{sin}(10\pi t))] \\ & = & \frac{∂}{∂t}[-50\pi \ \text{sin}(3\pi x)\text{sin}(4\pi y)\text{sin}(10\pi t)] \\ & = & -500{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t) \\ {u}_{xx} & = & \frac{∂}{∂x}[\frac{∂u}{∂x}] \\ & = & \frac{∂}{∂x}[15\pi \ \text{cos}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t)] \\ & = & -45{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t) \\ {u}_{yy} & = & \frac{∂}{∂y}[\frac{∂u}{∂y}] \\ & = & \frac{∂}{∂y}[5\ \text{sin}(3\pi x)(4\pi \ \text{cos}(4\pi y))\text{cos}(10\pi t)] \\ & = & \frac{∂}{∂y}[20\pi \ \text{sin}(3\pi x)\text{cos}(4\pi y)\text{cos}(10\pi t)] \\ & = & -80{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t).\end{array}\]

    Next, we substitute each of these into the right-hand side of and simplify:

    \[\begin{array}{ll}4({u}_{xx}+{u}_{yy}) & =4(-45{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t)+-80{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t)) \\ & =4(-125{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t)) \\ & =-500{\pi }^{2}\text{sin}(3\pi x)\text{sin}(4\pi y)\text{cos}(10\pi t) \\ & ={u}_{tt}.\end{array}\]

    This verifies the solution.

  14. Verify that \(u(x,y,t)=2\ \text{sin}(\frac{x}{3})\text{sin}(\frac{y}{4}){e}^{-25t\text{/}16}\) is a solution to the heat equation

    \[{u}_{t}=9({u}_{xx}+{u}_{yy}).\]

  15. \(\frac{∂z}{∂x}\) for \(z={x}^{2}-3xy+{y}^{2}\)

  16. \(\frac{∂z}{∂y}\) for \(z={x}^{2}-3xy+{y}^{2}\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    \(\frac{∂z}{∂y}=-3x+2y\)

  17. \({f}_{x}(1,1)\)

  18. \({f}_{x}(-1,1)\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    The sign is negative.

  19. \({f}_{y}(1,1)\)

  20. \({f}_{x}(0,0)\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    The partial derivative is zero at the origin.

  21. \(\frac{∂z}{∂x}\) for \(z=\text{sin}(3x)\text{cos}(3y)\)

  22. \(\frac{∂z}{∂y}\) for \(z=\text{sin}(3x)\text{cos}(3y)\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    \(\frac{∂z}{∂y}=-3\ \text{sin}(3x)\text{sin}(3y)\)

  23. \(\frac{∂z}{∂x}\) and \(\frac{∂z}{∂y}\) for \(z={x}^{8}{e}^{3y}\)

  24. \(\frac{∂z}{∂x}\) and \(\frac{∂z}{∂y}\) for \(z=\text{ln}({x}^{6}+{y}^{4})\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    \(\frac{∂z}{∂x}=\frac{6{x}^{5}}{{x}^{6}+{y}^{4}};\frac{∂z}{∂y}=\frac{4{y}^{3}}{{x}^{6}+{y}^{4}}\)

  25. Find \({f}_{y}(x,y)\) for \(f(x,y)={e}^{xy}\text{cos}(x)\text{sin}(y).\)

  26. Let \(z={e}^{xy}.\) Find \(\frac{∂z}{∂x}\) and \(\frac{∂z}{∂y}.\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    \(\frac{∂z}{∂x}=y{e}^{xy};\frac{∂z}{∂y}=x{e}^{xy}\)

  27. Let \(z=\text{ln}(\frac{x}{y}).\) Find \(\frac{∂z}{∂x}\) and \(\frac{∂z}{∂y}.\)

  28. Let \(z=\text{tan}(2x-y).\) Find \(\frac{∂z}{∂x}\) and \(\frac{∂z}{∂y}.\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    \(\frac{∂z}{∂x}=2\ {\text{sec}}^{2}(2x-y),\frac{∂z}{∂y}=\text{-}{\text{sec}}^{2}(2x-y)\)

  29. Let \(z=\text{sinh}(2x+3y).\) Find \(\frac{∂z}{∂x}\) and \(\frac{∂z}{∂y}.\)

  30. Let \(f(x,y)=\text{arctan}(\frac{y}{x}).\) Evaluate \({f}_{x}(2,-2)\) and \({f}_{y}(2,-2).\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    \({f}_{x}(2,-2)=\frac{1}{4}={f}_{y}(2,-2)\)

  31. Let \(f(x,y)=\frac{xy}{x-y}.\) Find \({f}_{x}(2,-2)\) and \({f}_{y}(2,-2).\)

  32. Find \(\frac{∂z}{∂x}\) at \((0,1)\) for \(z={e}^{\text{-}x}\text{cos}(y).\) Evaluate the partial derivatives at point \(P(0,1).\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    \(\frac{∂z}{∂x}=\text{-}\text{cos}(1)\)

  33. Given \(f(x,y,z)={x}^{3}y{z}^{2},\) find \(\frac{{∂}^{2}f}{∂x∂y}\) and \({f}_{z}(1,1,1).\)

  34. Given \(f(x,y,z)=2\ \text{sin}(x+y),\) find \({f}_{x}(0,\frac{\pi }{2},-4),\) \({f}_{y}(0,\frac{\pi }{2},-4),\) and \({f}_{z}(0,\frac{\pi }{2},-4).\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    \(\begin{array}{lll}{f}_{x}=0, & {f}_{y}=0, & {f}_{z}=0\end{array}\)

  35. The area of a parallelogram with adjacent side lengths that are \(a\ \text{and}\ b,\) and in which the angle between these two sides is \(\theta ,\) is given by the function \(A(a,b,\theta )=ba\ \text{sin}(\theta ).\) Find the rate of change of the area of the parallelogram with respect to the following:

    1. Side a
    2. Side b
    3. \(\text{Angle}\ \theta\)

  36. Express the volume of a right circular cylinder as a function of two variables:

    1. its radius \(r\) and its height \(h.\)
    2. Show that the rate of change of the volume of the cylinder with respect to its radius is the product of its circumference multiplied by its height.
    3. Show that the rate of change of the volume of the cylinder with respect to its height is equal to the area of the circular base.
    ഉത്തരം വെളിപ്പെടുത്തുക

    a. \(V(r,h)=\pi {r}^{2}h\) b. \(\frac{∂V}{∂r}=2\pi rh\) c. \(\frac{∂V}{∂h}=\pi {r}^{2}\)

  37. Calculate \(\frac{∂w}{∂z}\) for \(w=z\ \text{sin}(x{y}^{2}+2z).\)

  38. \({f}_{xy}\) for \(z=\text{ln}(x-y)\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    \({f}_{xy}=\frac{1}{{(x-y)}^{2}}\)

  39. \({f}_{yx}\) for \(z=\text{ln}(x-y)\)

  40. Let \(z={x}^{2}+3xy+2{y}^{2}.\) Find \(\frac{{∂}^{2}z}{∂{x}^{2}}\) and \(\frac{{∂}^{2}z}{∂{y}^{2}}.\)

    ഉത്തരം വെളിപ്പെടുത്തുക

    \(\frac{{∂}^{2}z}{∂{x}^{2}}=2,\frac{{∂}^{2}z}{∂{y}^{2}}=4\)

Symbols used here

\frac{\partial f}{\partial x}
partial derivative
Derivative with respect to x, holding the other variables fixed.
\nabla f
gradient (nabla, del)
Vector of partial derivatives; points uphill.
\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
\pi
pi
Ratio of a circle's circumference to its diameter, 3.14159…
A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}
matrix
A rectangular array of numbers; a linear map.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\int f(x)\,dx,\ \int_a^b
integral
Antiderivative (indefinite) or signed area from a to b (definite).
\iint,\ \oint
double / contour integral
Integral over a region of the plane; integral around a closed curve.
\mathbf{v},\ \vec{v}
vector
A quantity with magnitude and direction; a column of numbers.
\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|
dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Partial derivatives and the gradient

  1. To take ∂f/∂x, treat every other variable as a constant and differentiate as usual.
  2. Repeat for each variable.
  3. Assemble the results into the gradient vector.
  4. Set every component to zero to find critical points.

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

നീ സ്വയം ശ്രമിക്ക്.

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

കൂടുതല്‍ Multivariable Calculus

Functions of several variablesOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems