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Implicit differentiation

Differentiating relations that are not solved for y.

When y is defined implicitly by an equation in x and y, differentiate both sides treating y as a function of x (so y² becomes 2y y′), then solve for y′. Picture it: on the circle x² + y² = 25, the slope at (3, 4) is −x/y = −3/4, perpendicular to the radius. Think it: this is the chain rule applied to y(x), and the implicit function theorem says when it is legitimate.

Implicit Differentiation

In most discussions of math, if the dependent variable $y$ is a function of the independent variable $x,$ we express y in terms of $x.$ If this is the case, we say that $y$ is an explicit function of $x.$ For example, when we write the equation $y={x}^{2}+1,$ we are defining y explicitly in terms of $x.$ On the other hand, if the relationship between the function $y$ and the variable $x$ is expressed by an equation where $y$ is not expressed entirely in terms of $x,$ we say that the equation defines y implicitly in terms of $x.$ For example, the equation $y-{x}^{2}=1$ defines the function $y={x}^{2}+1$ implicitly.

Implicit differentiation allows us to find slopes of tangents to curves that are clearly not functions (they fail the vertical line test). We are using the idea that portions of $y$ are functions that satisfy the given equation, but that $y$ is not actually a function of $x.$

In general, an equation defines a function implicitly if the function satisfies that equation. An equation may define many different functions implicitly. For example, the functions

$y=\sqrt{25-{x}^{2}},y=-\sqrt{25-{x}^{2}},$ and y=\{\begin{array}{l}\sqrt{25-{x}^{2}}\ \text{if}-5

If we want to find the slope of the line tangent to the graph of ${x}^{2}+{y}^{2}=25$ at the point $(3,4),$ we could evaluate the derivative of the function $y=\sqrt{25-{x}^{2}}$ at $x=3.$ On the other hand, if we want the slope of the tangent line at the point $(3,-4),$ we could use the derivative of $y=\text{-}\sqrt{25-{x}^{2}}.$ However, it is not always easy to solve for a function defined implicitly by an equation. Fortunately, the technique of implicit differentiation allows us to find the derivative of an implicitly defined function without ever solving for the function explicitly. The process of finding $\frac{dy}{dx}$ using implicit differentiation is described in the following problem-solving strategy.

Condensed: the full section is in OpenStax Calculus Volume 1.

Finding Tangent Lines Implicitly

Now that we have seen the technique of implicit differentiation, we can apply it to the problem of finding equations of tangent lines to curves described by equations.

Example

Try it.

Find an equation of the line tangent to the curve ${x}^{2}+{y}^{2}=25$ at the point $(3,-4).$

Solution

Although we could find this equation without using implicit differentiation, using that method makes it much easier. In , we found $\frac{dy}{dx}=-\frac{x}{y}.$

The slope of the tangent line is found by substituting $(3,-4)$ into this expression. Consequently, the slope of the tangent line is $\frac{dy}{dx}|\begin{array}{l} \\ {}_{(3,-4)}\end{array}=-\frac{3}{-4}=\frac{3}{4}.$

Using the point $(3,-4)$ and the slope $\frac{3}{4}$ in the point-slope equation of the line, we obtain the equation $y=\frac{3}{4}x-\frac{25}{4}$ ().

Example

Try it.

Find an equation of the line tangent to the graph of ${y}^{3}+{x}^{3}-3xy=0$ at the point $(\frac{3}{2},\frac{3}{2})$ (). This curve is known as the folium (or leaf) of Descartes.

Solution

Begin by finding $\frac{dy}{dx}.$

\[\begin{array}{lll}\frac{d}{dx}({y}^{3}+{x}^{3}-3xy) & = & \frac{d}{dx}(0) \\ 3{y}^{2}\frac{dy}{dx}+3{x}^{2}-(3y+\frac{dy}{dx}3x) & = & 0 \\ \frac{dy}{dx} & = & \frac{3y-3{x}^{2}}{3{y}^{2}-3x}.\end{array}\]

Next, substitute $(\frac{3}{2},\frac{3}{2})$ into $\frac{dy}{dx}=\frac{3y-3{x}^{2}}{3{y}^{2}-3x}$ to find the slope of the tangent line:

\[\frac{dy}{dx}|\begin{array}{l} \\ {}_{(\frac{3}{2},\frac{3}{2})}\end{array}=-1.\]

Finally, substitute into the point-slope equation of the line to obtain

\[y=\text{-}x+3.\]

Example

Try it.

In a simple video game, a rocket travels in an elliptical orbit whose path is described by the equation $4{x}^{2}+25{y}^{2}=100.$ The rocket can fire missiles along lines tangent to its path. The object of the game is to destroy an incoming asteroid traveling along the positive x-axis toward $(0,0).$ If the rocket fires a missile when it is located at $(3,\frac{8}{5}),$ where will it intersect the x-axis?

Solution

To solve this problem, we must determine where the line tangent to the graph of

$4{x}^{2}+25{y}^{2}=100$ at $(3,\frac{8}{5})$ intersects the x-axis. Begin by finding $\frac{dy}{dx}$ implicitly.

Differentiating, we have

\[8x+50y\frac{dy}{dx}=0.\]

Solving for $\frac{dy}{dx},$ we have

\[\frac{dy}{dx}=-\frac{4x}{25y}.\]

The slope of the tangent line is $\frac{dy}{dx}|{}_{(3,\frac{8}{5})}=-\frac{3}{10}.$ The equation of the tangent line is $y=-\frac{3}{10}x+\frac{5}{2}.$ To determine where the line intersects the x-axis, solve $0=-\frac{3}{10}x+\frac{5}{2}.$ The solution is $x=\frac{25}{3}.$ The missile intersects the x-axis at the point $(\frac{25}{3},0).$

Condensed: the full section is in OpenStax Calculus Volume 1.

Key Concepts

We use implicit differentiation to find derivatives of implicitly defined functions (functions defined by equations).
By using implicit differentiation, we can find the equation of a tangent line to the graph of a curve.

Implicit Differentiation

For the following exercises, use implicit differentiation to find $\frac{dy}{dx}.$

For the following exercises, find an equation of the tangent line to the graph of the given equation at the indicated point. Use a calculator or computer software to graph the function and the tangent line.

For the following exercises, consider a closed rectangular box with a square base with side $x$ and height $y.$

For the following exercises, use implicit differentiation to determine ${y}^{'}.$ Does the answer agree with the formulas we have previously determined?

Implicit differentiation

In calculus, implicit differentiation is a method for finding the derivative of a function that is defined by an equation rather than by an explicit formula. If an equation such as

$F(x,y)=0$

defines $y$ as a function of $x$, at least locally, implicit differentiation treats $y$ as a function $y(x)$ and differentiates both sides of the equation with respect to $x$. The method is an application of the chain rule.

Implicit differentiation is useful when solving explicitly for one variable is inconvenient, produces several branches, or is not possible in elementary terms. For example, the circle $x^2+y^2=1$ cannot be represented globally as the graph of a single function $y=f(x)$, but its upper and lower arcs can each be differentiated implicitly.

The method allows for the computation of the tangent line approximation to $y(x)$, given only the function $F$ and the point $(x_0,y_0)$ at which the approximation is made, so that no other knowledge of the functional dependence of $y$ on $x$ is needed. Tangent approximations to surfaces and higher-dimensional manifolds given by an equation or system of equations can be found by similar methods.

Basic formula

Let $F$ be a differentiable function of two variables, and suppose that an equation

$F(x,y)=F(x_0,y_0)$

defines $y$ locally as a differentiable function of $x$, near the point $(x,y)=(x_0,y_0)$. Applying the differential to both sides gives \[dF = 0\] on the curve of interest, and writing the left-hand side in terms of the partial derivatives of $F$: \[F_x(x,y)dx + F_y(x,y)dy = 0.\] Therefore, evaluating at the point $(x_0,y_0)$ gives \[\frac{dy}{dx} = - \frac{F_x(x_0,y_0)}{F_y(x_0,y_0)}\] provided $F_y(x_0,y_0)\not=0$.

The derivative on the left-hand side of the formula can be interpreted as the slope of the tangent line to the curve defined implicitly by the equation $F(x,y)=F(x_0,y_0)$ through the given point $(x_0,y_0)$. The formula is local, in the sense that it gives only the derivative of the branch of the implicit function $y(x)$ whose graph contains the point $(x_0,y_0)$.

Relation with the implicit function theorem

Intuitively, in order to be valid at a point $(x_0,y_0)$ implicit differentiation in the basic formulation requires that the function $F$ must genuinely depend on $y$ near the point $(x_0,y_0)$, for otherwise the equation $F(x,y)=F(x_0,y_0)$ cannot be solved for $y$ as a function of $x$ at all.

More is actually required: it is clear from the formula that the partial derivative $F_y(x_0,y_0)$ must be non-zero, because it appears in the denominator of the right-hand side of the formula for $dy/dx_{(x_0,y_0)}$. While this fact allows one to compute a numerical value, it is not obvious that the numerical value has the advertised meaning: namely, that it be the derivative of a function $y(x)$ whose graph is contained on the curve $F(x,y)=0$ through the point.

The implicit function theorem supplies the missing justification. It asserts as follows: suppose that $F(x,y)$ is a function such that both partial derivatives exist in a disc around $(x,y)=(x_0,y_0)$, and are continuous throughout the disc. Then if $F_y(x_0,y_0) \ne 0$, there is a differentiable function $y(x)$ defined in an open interval containing $x=x_0$, such that $F(x,y(x))=0$ throughout the interval, and whose derivative is given by the implicit differentiation formula at $(x_0,y_0)$. The function is, moreover, unique on a sufficiently small interval around $x_0$.

Example 1

Consider

$y + x + 5 = 0 \,.$

This equation is easy to solve for y, giving

$y = -x - 5 \,,$

where the right side is the explicit form of the function y(x). Differentiation then gives ⁠dy/dx⁠ = −1.

Alternatively, one can totally differentiate the original equation:

$\begin{aligned} \frac{dy}{dx} + \frac{dx}{dx} + \frac{d}{dx}(5) &= 0 \, ; \\[6px] \frac{dy}{dx} + 1 + 0 &= 0 \,. \end{aligned}$

Solving for ⁠dy/dx⁠ gives

$\frac{dy}{dx} = -1 \,,$

the same answer as obtained previously.

Example 2

An example of an implicit function for which implicit differentiation is easier than using explicit differentiation is the function y(x) defined by the equation

$x^4 + 2y^2 = 8 \,.$

To differentiate this explicitly with respect to x, one has first to get

$y(x) = \pm\sqrt{\frac{8 - x^4}{2}} \,,$

and then differentiate this function. This creates two derivatives: one for y ≥ 0 and another for y < 0.

It is substantially easier to implicitly differentiate the original equation:

$4x^3 + 4y\frac{dy}{dx} = 0 \,,$

giving

$\frac{dy}{dx} = \frac{-4x^3}{4y} = -\frac{x^3}{y} \,.$

Example 3

Often, it is difficult or impossible to solve explicitly for y, and implicit differentiation is the only feasible method of differentiation. An example is the equation

$y^5-y=x \,.$

It is impossible to express y explicitly in radicals as a function of x, and therefore one cannot find ⁠dy/dx⁠ by explicit differentiation of elementary radical expressions. Using the implicit method, ⁠dy/dx⁠ can be obtained by differentiating the equation to obtain

$5y^4\frac{dy}{dx} - \frac{dy}{dx} = \frac{dx}{dx} \,,$

where ⁠dx/dx⁠ = 1. Factoring out ⁠dy/dx⁠ shows that

$\left(5y^4 - 1\right)\frac{dy}{dx} = 1 \,,$

which yields the result

$\frac{dy}{dx}=\frac{1}{5y^4-1} \,,$

which is defined for

$y \ne \pm\frac{1}{\sqrt[4]{5}} \quad \text{and} \quad y \ne \pm \frac{i}{\sqrt[4]{5}} \,.$

Higher order derivatives

The method of implicit differentiation can be applied to find higher-order derivatives such as the second derivative of the implicitly-defined function $y(x)$ at the point of its graph $(x_0,y_0)$, provided the function $F$ is sufficiently smooth at the point.

For example, suppose that the first derivative of $y(x)$ has been found, by solving the equation \[dF = F_x(x_0,y_0)\,dx + F_y(x_0,y_0)\,dy = 0\] for $dy/dx$ at $(x_0,y_0)$.

Applying the second differential to $F(x,y) = F(x_0,y_0)$ then gives \[d^2 F = 0\] and expanding the left-hand side gives \[d(F_x\,dx + F_y\,dy) = F_{xx}dx^2 + F_{yx}dx\,dy + F_{xy}dy\,dx + F_{yy}dy^2 + F_x\,d^2x + F_y\,d^2y=0.\] where we have written the second partial derivatives of $F$ with repeated subscripts. Since it is the $x$ derivatives we are after, $d^2x=0$. Also, substituting the known expressions for $dy$ at the point $(x_0,y_0)$, into the formula (that is $dy = \frac{dy}{dx}dx$ where the derivative on the right-hand side is known at the point $(x_0,y_0)$: $dy/dx_{(x_0,y_0)} = -F_x(x_0,y_0)/F_y(x_0,y_0)$), we can then arrange terms and solve for the second derivative $d^2y/dx^2$. That is, from \[F_{xx} dx^2 + (F_{xy}+F_{yx})\frac{dy}{dx}dx^2 + F_{yy}dy^2 + F_y\,d^2y= 0\] we obtain \[\frac{d^2y}{dx^2} = - \frac { F_{xx} + (F_{xy}+F_{yx})\frac{dy}{dx}+ F_{yy}\left(\frac{dy}{dx}\right)^2 } { F_{y} } .\] This is understood as at the point $(x_0,y_0)$, and requires only that the function $F$ be twice-differentiable at the point, in addition to the hypothesis $F_y(x_0,y_0)\ne 0$. Second-differentiability at the point is true if the first and second partials of $F$ exist and are continuous in a disc around the point.

In that case, Clairaut's theorem implies that the mixed partials are equal $F_{xy}(x_0,y_0)=F_{yx}(x_0,y_0)$, and these terms can then be combined to give \[\frac{d^2y}{dx^2} = - \frac { F_{xx} - 2F_{xy}\frac{F_x}{F_y} + F_{yy}\left(\frac{F_x}{F_y}\right)^2 } { F_{y} } .\]

Higher order derivatives (under the requisite smoothness hypotheses) are handled similarly.

Worked example · derivative of x^2 + y^2 wrt x

Calculus · Derivatives

Differentiate x^2 + y^2

\frac{d}{dx}\left[x^{2} + y^{2}\right]

Step by step

\frac{d}{dx}\left[x^{2} + y^{2}\right]
Start from the derivative to compute.
\frac{d}{d x} x^{2} + \frac{d}{d x} y^{2}
Sum rule: differentiate term by term.
2 x + \frac{d}{d x} y^{2}
Power rule: (xⁿ)′ = n·xⁿ⁻¹ with n = 2.
2 x
The derivative of a constant is 0.

Reveal the answer

f'(x) = 2 x

Now you Pick a problem, or type or draw your own. Every step, a picture, the answer hidden until you ask.

Practice (39)

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

Assuming that $y$ is defined implicitly by the equation ${x}^{2}+{y}^{2}=25,$ find $\frac{dy}{dx}.$

Reveal the answer

Follow the steps in the problem-solving strategy.
\[\begin{array}{llllll}\frac{d}{dx}({x}^{2}+{y}^{2}) & = & \frac{d}{dx}(25) & & & \text{Step 1. Differentiate both sides of the equation.} \\ \frac{d}{dx}({x}^{2})+\frac{d}{dx}({y}^{2}) & = & 0 & & & \begin{array}{l}\text{Step 1.1. Use the sum rule on the left.} \\ \text{On the right}\ \frac{d}{dx}(25)=0.\end{array} \\ 2x+2y\frac{dy}{dx} & = & 0 & & & \begin{array}{l}\text{Step 1.2. Take the derivatives, so}\ \frac{d}{dx}({x}^{2})=2x \\ \text{and}\ \frac{d}{dx}({y}^{2})=2y\frac{dy}{dx}.\end{array} \\ 2y\frac{dy}{dx} & = & -2x & & & \begin{array}{l}\text{Step 2. Keep the terms with}\ \frac{dy}{dx}\ \text{on the left.} \\ \text{Move the remaining terms to the right.}\end{array} \\ \frac{dy}{dx} & = & -\frac{x}{y} & & & \begin{array}{l}\text{Step 4. Divide both sides of the equation by} \\ 2y.\ \text{(Step 3 does not apply in this case.)}\end{array}\end{array}\]
Assuming that $y$ is defined implicitly by the equation ${x}^{3}\ \text{sin}\ y+y=4x+3,$ find $\frac{dy}{dx}.$

Reveal the answer

\[\begin{array}{llllll}\frac{d}{dx}({x}^{3}\text{sin}\ y+y) & = & \frac{d}{dx}(4x+3) & & & \text{Step 1: Differentiate both sides of the equation.} \\ \frac{d}{dx}({x}^{3}\text{sin}\ y)+\frac{d}{dx}(y) & = & 4 & & & \begin{array}{l}\text{Step 1.1: Apply the sum rule on the left.} \\ \text{On the right,}\ \frac{d}{dx}(4x+3)=4.\end{array} \\ (\frac{d}{dx}({x}^{3})\cdot \text{sin}\ y+\frac{d}{dx}(\text{sin}\ y)\cdot {x}^{3})+\frac{dy}{dx} & = & 4 & & & \begin{array}{l}\text{Step 1.2: Use the product rule to find} \\ \frac{d}{dx}({x}^{3}\text{sin}\ y).\ \text{Observe that}\ \frac{d}{dx}(y)=\frac{dy}{dx}.\end{array} \\ 3{x}^{2}\text{sin}\ y+(\text{cos}\ y\frac{dy}{dx})\cdot {x}^{3}+\frac{dy}{dx} & = & 4 & & & \begin{array}{l}\text{Step 1.3: We know}\ \frac{d}{dx}({x}^{3})=3{x}^{2}.\ \text{Use the} \\ \text{chain rule to obtain}\ \frac{d}{dx}(\text{sin}\ y)=\text{cos}\ y\frac{dy}{dx}.\end{array} \\ {\text{x}}^{3}\text{cos}\ y\frac{dy}{dx}+\frac{dy}{dx} & = & 4-3{x}^{2}\text{sin}\ y & & & \begin{array}{l}\text{Step 2: Keep all terms containing}\ \frac{dy}{dx}\ \text{on the} \\ \text{left. Move all other terms to the right.}\end{array} \\ \frac{dy}{dx}({\text{x}}^{3}\text{cos}\ y+1) & = & 4-3{x}^{2}\text{sin}\ y & & & \text{Step 3: Factor out}\ \frac{dy}{dx}\ \text{on the left.} \\ \frac{dy}{dx} & = & \frac{4-3{x}^{2}\text{sin}\ y}{{x}^{3}\text{cos}\ y+1} & & & \begin{array}{l}\text{Step 4: Solve for}\ \frac{dy}{dx}\ \text{by dividing both sides of} \\ \text{the equation by}\ {\text{x}}^{3}\text{cos}\ y+1.\end{array}\end{array}\]
Find $\frac{{d}^{2}y}{d{x}^{2}}$ if ${x}^{2}+{y}^{2}=25.$

Reveal the answer

In , we showed that $\frac{dy}{dx}=-\frac{x}{y}.$ We can take the derivative of both sides of this equation to find $\frac{{d}^{2}y}{d{x}^{2}}.$
\[\begin{array}{lllll}\frac{{d}^{2}y}{d{x}^{2}} & =\frac{d}{dx}(-\frac{x}{y}) & & & \text{Differentiate both sides of}\ \frac{dy}{dx}=-\frac{x}{y}. \\ & =-\frac{(1\cdot y-x\frac{dy}{dx})}{{y}^{2}} & & & \text{Use the quotient rule to find}\ \frac{d}{dy}(-\frac{x}{y}). \\ & =\frac{\text{-}y+x\frac{dy}{dx}}{{y}^{2}} & & & \text{Simplify.} \\ & =\frac{\text{-}y+x(-\frac{x}{y})}{{y}^{2}} & & & \text{Substitute}\ \frac{dy}{dx}=-\frac{x}{y}. \\ & =\frac{\text{-}{y}^{2}-{x}^{2}}{{y}^{3}} & & & \text{Simplify.}\end{array}\]
At this point we have found an expression for $\frac{{d}^{2}y}{d{x}^{2}}.$ If we choose, we can simplify the expression further by recalling that ${x}^{2}+{y}^{2}=25$ and making this substitution in the numerator to obtain $\frac{{d}^{2}y}{d{x}^{2}}=-\frac{25}{{y}^{3}}.$
Find $\frac{dy}{dx}$ for $y$ defined implicitly by the equation $4{x}^{5}+\text{tan}\ y={y}^{2}+5x.$

Reveal the answer

$\frac{dy}{dx}=\frac{5-20{x}^{4}}{{\text{sec}}^{2}y-2y}$
Find an equation of the line tangent to the curve ${x}^{2}+{y}^{2}=25$ at the point $(3,-4).$

Reveal the answer

Although we could find this equation without using implicit differentiation, using that method makes it much easier. In , we found $\frac{dy}{dx}=-\frac{x}{y}.$
The slope of the tangent line is found by substituting $(3,-4)$ into this expression. Consequently, the slope of the tangent line is $\frac{dy}{dx}|\begin{array}{l} \\ {}_{(3,-4)}\end{array}=-\frac{3}{-4}=\frac{3}{4}.$
Using the point $(3,-4)$ and the slope $\frac{3}{4}$ in the point-slope equation of the line, we obtain the equation $y=\frac{3}{4}x-\frac{25}{4}$ ().
Find an equation of the line tangent to the graph of ${y}^{3}+{x}^{3}-3xy=0$ at the point $(\frac{3}{2},\frac{3}{2})$ (). This curve is known as the folium (or leaf) of Descartes.

Reveal the answer

Begin by finding $\frac{dy}{dx}.$
\[\begin{array}{lll}\frac{d}{dx}({y}^{3}+{x}^{3}-3xy) & = & \frac{d}{dx}(0) \\ 3{y}^{2}\frac{dy}{dx}+3{x}^{2}-(3y+\frac{dy}{dx}3x) & = & 0 \\ \frac{dy}{dx} & = & \frac{3y-3{x}^{2}}{3{y}^{2}-3x}.\end{array}\]
Next, substitute $(\frac{3}{2},\frac{3}{2})$ into $\frac{dy}{dx}=\frac{3y-3{x}^{2}}{3{y}^{2}-3x}$ to find the slope of the tangent line:
\[\frac{dy}{dx}|\begin{array}{l} \\ {}_{(\frac{3}{2},\frac{3}{2})}\end{array}=-1.\]
Finally, substitute into the point-slope equation of the line to obtain
\[y=\text{-}x+3.\]
In a simple video game, a rocket travels in an elliptical orbit whose path is described by the equation $4{x}^{2}+25{y}^{2}=100.$ The rocket can fire missiles along lines tangent to its path. The object of the game is to destroy an incoming asteroid traveling along the positive x-axis toward $(0,0).$ If the rocket fires a missile when it is located at $(3,\frac{8}{5}),$ where will it intersect the x-axis?

Reveal the answer

To solve this problem, we must determine where the line tangent to the graph of
$4{x}^{2}+25{y}^{2}=100$ at $(3,\frac{8}{5})$ intersects the x-axis. Begin by finding $\frac{dy}{dx}$ implicitly.
Differentiating, we have
\[8x+50y\frac{dy}{dx}=0.\]
Solving for $\frac{dy}{dx},$ we have
\[\frac{dy}{dx}=-\frac{4x}{25y}.\]
The slope of the tangent line is $\frac{dy}{dx}|{}_{(3,\frac{8}{5})}=-\frac{3}{10}.$ The equation of the tangent line is $y=-\frac{3}{10}x+\frac{5}{2}.$ To determine where the line intersects the x-axis, solve $0=-\frac{3}{10}x+\frac{5}{2}.$ The solution is $x=\frac{25}{3}.$ The missile intersects the x-axis at the point $(\frac{25}{3},0).$
Find an equation of the line tangent to the hyperbola ${x}^{2}-{y}^{2}=16$ at the point $(5,3).$

Reveal the answer

$y=\frac{5}{3}x-\frac{16}{3}$
${x}^{2}-{y}^{2}=4$
$6{x}^{2}+3{y}^{2}=12$

Reveal the answer

$\frac{dy}{dx}=\frac{-2x}{y}$
${x}^{2}y=y-7$
$3{x}^{3}+9x{y}^{2}=5{x}^{3}$

Reveal the answer

$\frac{dy}{dx}=\frac{x}{3y}-\frac{y}{2x}$
$xy-\text{cos}\ (xy)=1$
$y\sqrt{x+4}=xy+8$

Reveal the answer

$\frac{dy}{dx}=\frac{y-\frac{y}{2\sqrt{x+4}}}{\sqrt{x+4}-x}$
$\text{-}xy-2=\frac{x}{7}$
$y\ \text{sin}\ (xy)={y}^{2}-2$

Reveal the answer

$\frac{dy}{dx}=\frac{{y}^{2}\text{cos}\ (xy)}{2y-\text{sin}\ (xy)-xy\ \text{cos}\ (xy)}$
${(xy)}^{2}+3x={y}^{2}$
${x}^{3}y+x{y}^{3}=-8$

Reveal the answer

$\frac{dy}{dx}=\frac{-3{x}^{2}y-{y}^{3}}{{x}^{3}+3x{y}^{2}}$
[T] ${x}^{4}y-x{y}^{3}=-2,(-1,-1)$
[T] ${x}^{2}{y}^{2}+5xy=14,(2,1)$

Reveal the answer

$y=\frac{-1}{2}x+2$
[T] $\text{tan}\ (xy)=y,(\frac{\pi }{4},1)$
[T] $x{y}^{2}+\text{sin}\ (\pi y)-2{x}^{2}=10,(2,-3)$

Reveal the answer

$y=\frac{1}{\pi +12}x-\frac{3\pi +38}{\pi +12}$
[T] $\frac{x}{y}+5x-7=-\frac{3}{4}y,(1,2)$
[T] $xy+\text{sin}\ (x)=1,(\frac{\pi }{2},0)$

Reveal the answer

$y=0$
[T] The graph of a folium of Descartes with equation $2{x}^{3}+2{y}^{3}-9xy=0$ is given in the following graph.
1. Find an equation of the tangent line at the point $(2,1).$ Graph the tangent line along with the folium.
2. Find an equation of the normal line to the tangent line in a. at the point $(2,1).$
For the equation ${x}^{2}+2xy-3{y}^{2}=0,$
1. Find an equation of the normal to the tangent line at the point $(1,1).$
2. At what other point does the normal line in a. intersect the graph of the equation?
Reveal the answer

a. $y=\text{-}x+2$ b. $(3,-1)$
Find all points on the graph of ${y}^{3}-27y={x}^{2}-90$ at which the tangent line is vertical.
For the equation ${x}^{2}+xy+{y}^{2}=7,$
1. Find the $x$-intercept(s).
2. Find the slope of the tangent line(s) at the x-intercept(s).
3. What does the value(s) in b. indicate about the tangent line(s)?
Reveal the answer

a. $(\pm \sqrt{7},0)$ b. $-2$ c. They are parallel since the slope is the same at both intercepts.
Find an equation of the tangent line to the graph of the equation ${\text{sin}}^{-1}x+{\text{sin}}^{-1}y=\frac{\pi }{6}$ at the point $(0,\frac{1}{2}).$
Find an equation of the tangent line to the graph of the equation ${\text{tan}}^{-1}(x+y)={x}^{2}+\frac{\pi }{4}$ at the point $(0,1).$

Reveal the answer

$y=\text{-}x+1$
Find ${y}^{'}$ and ${y}^{″}$ for ${x}^{2}+6xy-2{y}^{2}=3.$
[T] The number of cell phones produced when $x$ dollars is spent on labor and $y$ dollars is spent on capital invested by a manufacturer can be modeled by the equation $60{x}^{3\text{/}4}{y}^{1\text{/}4}=3240.$
1. Find $\frac{dy}{dx}$ and evaluate at the point $(81,16).$
2. Interpret the result of a.
Reveal the answer

a. $-0.5926$ b. When $81 is spent on labor and $16 is spent on capital, the amount spent on capital is decreasing by $0.5926 per $1 spent on labor.
[T] The number of cars produced when $x$ dollars is spent on labor and $y$ dollars is spent on capital invested by a manufacturer can be modeled by the equation $30{x}^{1\text{/}3}{y}^{2\text{/}3}=360.$
(Both $x$ and $y$ are measured in thousands of dollars.)
1. Find $\frac{dy}{dx}$ and evaluate at the point $(27,8).$
2. Interpret the result of a.
The volume of a right circular cone of radius $x$ and height $y$ is given by $V=\frac{1}{3}\pi {x}^{2}y.$ Suppose that the volume of the cone is a constant. Find $\frac{dy}{dx}$ when $x=4$ and $y=16.$

Reveal the answer

$-8$
Find an equation for the surface area of the rectangular box, $S(x,y).$
If the surface area of the rectangular box is 78 square feet, find $\frac{dy}{dx}$ when $x=3$ feet and $y=5$ feet.

Reveal the answer

$-2.67$
$x=\text{sin}\ y$
$x=\text{cos}\ y$

Reveal the answer

${y}^{'}=-\frac{1}{\sqrt{1-{x}^{2}}}$
$x=\text{tan}\ y$

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How to: Implicit differentiation

Find the derivative of a complicated function by using implicit differentiation.
Use implicit differentiation to determine the equation of a tangent line.
Take the derivative of both sides of the equation. Keep in mind that
Rewrite the equation so that all terms containing
Factor out
Solve for
We use implicit differentiation to find derivatives of implicitly defined functions (functions defined by equations).
By using implicit differentiation, we can find the equation of a tangent line to the graph of a curve.

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point, the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate: the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function, a polynomial times an exponential, log or trig function.

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), Wikipedia (CC BY-SA 4.0). Condensed and re-explained here; errors are ours. Because its source is licensed for non-commercial use, this lesson is free, sells nothing, and is shared under the same licence.

More in Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests