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Differentiation rules
Unless otherwise stated, all functions are functions of real numbers () that return real values, although, more generally, the formulas below apply wherever they are well defined, including the case of complex numbers…
The Basic Rules
The functions \(f(x)=c\) and \(g(x)={x}^{n}\) where \(n\) is a positive integer are the building blocks from which all polynomials and rational functions are constructed. To find derivatives of polynomials and rational functions efficiently without resorting to the limit definition of the derivative, we must first develop formulas for differentiating these basic functions.
We first apply the limit definition of the derivative to find the derivative of the constant function, \(f(x)=c.\) For this function, both \(f(x)=c\) and \(f(x+h)=c,\) so we obtain the following result:
\[\begin{array}{ll}{f}^{'}(x) & =\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h} \\ & =\underset{h\to 0}{\text{lim}}\frac{c-c}{h} \\ & =\underset{h\to 0}{\text{lim}}\frac{0}{h} \\ & =\underset{h\to 0}{\text{lim}}0=0.\end{array}\]The rule for differentiating constant functions is called the constant rule. It states that the derivative of a constant function is zero; that is, since a constant function is a horizontal line, the slope, or the rate of change, of a constant function is \(0.\) We restate this rule in the following theorem.
Example
Try it.
Find the derivative of \(f(x)=8.\)
Solution
This is just a one-step application of the rule:
\[{f}^{'}(x)=0.\]The Power Rule
We have shown that
\[\frac{d}{dx}({x}^{2})=2x\ \text{and}\ \frac{d}{dx}({x}^{1\text{/}2})=\frac{1}{2}{x}^{\text{-}1\text{/}2}.\]At this point, you might see a pattern beginning to develop for derivatives of the form \(\frac{d}{dx}({x}^{n}).\) We continue our examination of derivative formulas by differentiating power functions of the form \(f(x)={x}^{n}\) where \(n\) is a positive integer. We develop formulas for derivatives of this type of function in stages, beginning with positive integer powers. Before stating and proving the general rule for derivatives of functions of this form, we take a look at a specific case, \(\frac{d}{dx}({x}^{3}).\) As we go through this derivation, note that the technique used in this case is essentially the same as the technique used to prove the general case.
Example
Try it.
Find \(\frac{d}{dx}({x}^{3}).\)
Solution
\[\begin{array}{lllll}\frac{d}{dx}({x}^{3}) & =\underset{h\to 0}{\text{lim}}\frac{{(x+h)}^{3}-{x}^{3}}{h} & & & \\ & =\underset{h\to 0}{\text{lim}}\frac{{x}^{3}+3{x}^{2}h+3x{h}^{2}+{h}^{3}-{x}^{3}}{h} & & & \begin{array}{l}\text{Notice that the first term in the expansion of} \\ {(x+h)}^{3}\ \text{is}\ {x}^{3}\ \text{and the second term is}\ 3{x}^{2}h.\ \text{All} \\ \text{other terms contain powers of}\ h\ \text{that are two or} \\ \text{greater.}\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{3{x}^{2}h+3x{h}^{2}+{h}^{3}}{h} & & & \begin{array}{l}\text{In this step the}\ {x}^{3}\ \text{terms have been cancelled,} \\ \text{leaving only terms containing}\ h.\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{h(3{x}^{2}+3xh+{h}^{2})}{h} & & & \text{Factor out the common factor of}\ h. \\ & =\underset{h\to 0}{\text{lim}}(3{x}^{2}+3xh+{h}^{2}) & & & \begin{array}{l}\text{After cancelling the common factor of}\ h,\text{the} \\ \text{only term not containing}\ h\ \text{is}\ 3{x}^{2}.\end{array} \\ & =3{x}^{2} & & & \text{Let}\ h\ \text{go to 0.}\end{array}\]Condensed: the full section is in OpenStax Calculus Volume 1.
The Sum, Difference, and Constant Multiple Rules
We find our next differentiation rules by looking at derivatives of sums, differences, and constant multiples of functions. Just as when we work with functions, there are rules that make it easier to find derivatives of functions that we add, subtract, or multiply by a constant. These rules are summarized in the following theorem.
Condensed: the full section is in OpenStax Calculus Volume 1.
The Product Rule
Now that we have examined the basic rules, we can begin looking at some of the more advanced rules. The first one examines the derivative of the product of two functions. Although it might be tempting to assume that the derivative of the product is the product of the derivatives, similar to the sum and difference rules, the product rule does not follow this pattern. To see why we cannot use this pattern, consider the function \(f(x)={x}^{2},\) whose derivative is \({f}^{'}(x)=2x\) and not \(\frac{d}{dx}(x)\cdot \frac{d}{dx}(x)=1\cdot 1=1.\)
Condensed: the full section is in OpenStax Calculus Volume 1.
The Quotient Rule
Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the function in the numerator times the function in the denominator minus the derivative of the function in the denominator times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp why we cannot simply take the quotient of the derivatives, keep in mind that
\[\frac{d}{dx}({x}^{2})=2x,\text{not}\ \frac{\frac{d}{dx}({x}^{3})}{\frac{d}{dx}(x)}=\frac{3{x}^{2}}{1}=3{x}^{2}.\]The proof of the quotient rule is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example.
Example
Try it.
Use the quotient rule to find the derivative of \(k(x)=\frac{5{x}^{2}}{4x+3}.\)
Solution
Let \(f(x)=5{x}^{2}\) and \(g(x)=4x+3.\) Thus, \({f}^{'}(x)=10x\) and \({g}^{'}(x)=4.\) Substituting into the quotient rule, we have
\[{k}^{'}(x)=\frac{{f}^{'}(x)g(x)-{g}^{'}(x)f(x)}{{(g(x))}^{2}}=\frac{10x(4x+3)-4(5{x}^{2})}{{(4x+3)}^{2}}.\]Simplifying, we obtain
\[{k}^{'}(x)=\frac{20{x}^{2}+30x}{{(4x+3)}^{2}}.\]It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form \({x}^{k}\) where \(k\) is a negative integer.
Condensed: the full section is in OpenStax Calculus Volume 1.
Combining Differentiation Rules
As we have seen throughout the examples in this section, it seldom happens that we are called on to apply just one differentiation rule to find the derivative of a given function. At this point, by combining the differentiation rules, we may find the derivatives of any polynomial or rational function. Later on we will encounter more complex combinations of differentiation rules. A good rule of thumb to use when applying several rules is to apply the rules in reverse of the order in which we would evaluate the function.
Example
Try it.
For \(k(x)=3h(x)+{x}^{2}g(x),\) find \({k}^{'}(x).\)
Solution
Finding this derivative requires the sum rule, the constant multiple rule, and the product rule.
\[\begin{array}{lllll}{k}^{'}(x) & =\frac{d}{dx}(3h(x)+{x}^{2}g(x))=\frac{d}{dx}(3h(x))+\frac{d}{dx}({x}^{2}g(x)) & & & \text{Apply the sum rule.} \\ & =3\frac{d}{dx}(h(x))+(\frac{d}{dx}({x}^{2})g(x)+\frac{d}{dx}(g(x)){x}^{2}) & & & \begin{array}{l}\text{Apply the constant multiple rule to} \\ \text{differentiate}\ 3h(x)\ \text{and the product} \\ \text{rule to differentiate}\ {x}^{2}g(x).\end{array} \\ & =3{h}^{'}(x)+2xg(x)+{x}^{2}{g}^{'}(x) & & & \end{array}\]Example
Try it.
For \(k(x)=f(x)g(x)h(x),\) express \({k}^{'}(x)\) in terms of \(f(x),g(x),h(x),\) and their derivatives.
Solution
We can think of the function \(k(x)\) as the product of the function \(f(x)g(x)\) and the function \(h(x).\) That is, \(k(x)=(f(x)g(x))\cdot h(x).\) Thus,
\[\begin{array}{lllll}{k}^{'}(x) & =\frac{d}{dx}(f(x)g(x))\cdot h(x)+\frac{d}{dx}(h(x))\cdot (f(x)g(x)) & & & \begin{array}{l}\text{Apply the product rule to the product} \\ \text{of}\ f(x)g(x)\ \text{and}\ h(x).\end{array} \\ & =({f}^{'}(x)g(x)+{g}^{'}(x)f(x))h(x)+{h}^{'}(x)f(x)g(x) & & & \text{Apply the product rule to}\ f(x)g(x). \\ & ={f}^{'}(x)g(x)h(x)+f(x){g}^{'}(x)h(x)+f(x)g(x){h}^{'}(x). & & & \text{Simplify.}\end{array}\]Example
Try it.
Determine the values of \(x\) for which \(f(x)={x}^{3}-7{x}^{2}+8x+1\) has a horizontal tangent line.
Solution
To find the values of \(x\) for which \(f(x)\) has a horizontal tangent line, we must solve \({f}^{'}(x)=0.\) Since
\[{f}^{'}(x)=3{x}^{2}-14x+8=(3x-2)(x-4),\]we must solve \((3x-2)(x-4)=0.\) Thus we see that the function has horizontal tangent lines at \(x=\frac{2}{3}\) and \(x=4\) as shown in the following graph.
Condensed: the full section is in OpenStax Calculus Volume 1.
Key Concepts
- The derivative of a constant function is zero.
- The derivative of a power function is a function in which the power on \(x\) becomes the coefficient of the term and the power on \(x\) in the derivative decreases by 1.
- The derivative of a constant c multiplied by a function f is the same as the constant multiplied by the derivative.
- The derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and the derivative of g.
- The derivative of the difference of a function f and a function g is the same as the difference of the derivative of f and the derivative of g.
- The derivative of a product of two functions is the derivative of the first function times the second function plus the derivative of the second function times the first function.
- The derivative of the quotient of two functions is the derivative of the first function times the second function minus the derivative of the second function times the first function, all divided by the square of the second function.
- We used the limit definition of the derivative to develop formulas that allow us to find derivatives without resorting to the definition of the derivative. These formulas can be used singly or in combination with each other.
Differentiation Rules
For the following exercises, find \({f}^{'}(x)\) for each function.
For the following exercises, find an equation of the tangent line \(T(x)\) to the graph of the given function at the indicated point. Use a graphing calculator to graph the function and the tangent line.
For the following exercises, assume that \(f(x)\) and \(g(x)\) are both differentiable functions for all \(x.\) Find the derivative of each of the functions \(h(x).\)
For the following exercises, assume that \(f(x)\) and \(g(x)\) are both differentiable functions with values as given in the following table. Use the following table to calculate the following derivatives.
| \(x\) | \(1\) | \(2\) | \(3\) | \(4\) |
| \(f(x)\) | \(3\) | \(5\) | \(-2\) | \(0\) |
| \(g(x)\) | \(2\) | \(3\) | \(-4\) | \(6\) |
| \({f}^{'}(x)\) | \(-1\) | \(7\) | \(8\) | \(-3\) |
| \({g}^{'}(x)\) | \(4\) | \(1\) | \(2\) | \(9\) |
For the following exercises, use the following figure to find the indicated derivatives, if they exist.
For the following exercises,
- evaluate \({f}^{'}(a),\) and
- graph the function \(f(x)\) and the tangent line at \(x=a.\)
Elementary rules of differentiation
Unless otherwise stated, all functions are functions of real numbers (\(\mathbb{R}\)) that return real values, although, more generally, the formulas below apply wherever they are well defined, including the case of complex numbers (\(\mathbb{C}\)).
Differentiation rules
Let \(c \in \mathbb{R}\) and \(f(x) = c\). By the definition of the derivative: \[\begin{aligned} f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{(c) - (c)}{h} \\ &= \lim_{h \to 0} \frac{0}{h} \\ &= \lim_{h \to 0} 0 \\ &= 0. \end{aligned}\]
This computation shows that the derivative of any constant function is 0.
Linearity of differentiation
For any functions \(f\) and \(g\) and any real numbers \(a\) and \(b\), the derivative of the function \(h(x) = af(x) + bg(x)\) with respect to \(x\) is \(h'(x) = a f'(x) + b g'(x)\).
In Leibniz's notation, this formula is written as: \[\frac{d(af+bg)}{dx} = a\frac{df}{dx} +b\frac{dg}{dx}.\]
Special cases include:
- The constant factor rule:
\[(af)' = af',\]
- The sum rule:
\[(f + g)' = f' + g',\]
- The difference rule:
\[(f - g)' = f' - g'.\]
Product rule
For the functions \(f\) and \(g\), the derivative of the function \(h(x) = f(x) g(x)\) with respect to \(x\) is: \[h'(x) = (fg)'(x) = f'(x) g(x) + f(x) g'(x).\]
In Leibniz's notation, this formula is written: \[\frac{d(fg)}{dx} = g \frac{df}{dx} + f \frac{dg}{dx}.\]
Chain rule
The derivative of the function \(h(x) = f(g(x))\) is: \[h'(x) = f'(g(x))\cdot g'(x).\]
In Leibniz's notation, this formula is written as: \[\frac{d}{dx}h(x) = \left.\frac{d}{dz}f(z)\right|_{z=g(x)}\cdot \frac{d}{dx}g(x),\] often abridged to: \[\frac{dh(x)}{dx} = \frac{df(g(x))}{dg(x)} \cdot \frac{dg(x)}{dx}.\]
Focusing on the notion of maps, and the differential being a map \(\text{D}\), this formula is written in a more concise way as: \[[\text{D} (f\circ g)]_x = [\text{D} f]_{g(x)} \cdot [\text{D}g]_x.\]
Inverse function rule
If the function \(f\) has an inverse function \(g\), meaning that \(g(f(x)) = x\) and \(f(g(y)) = y\), then: \[g' = \frac{1}{f'\circ g}.\]
In Leibniz notation, this formula is written as: \[\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}.\]
Polynomial or elementary power rule
If \(f(x) = x^r\), for any real number \(r \neq 0\), then: \[f'(x) = rx^{r-1}.\]
When \(r = 1\), this formula becomes the special case that, if \(f(x) = x\), then \(f'(x) = 1\).
Combining the power rule with the sum and constant multiple rules permits the computation of the derivative of any polynomial.
Reciprocal rule
The derivative of \(h(x)=\frac{1}{f(x)}\) for any (nonvanishing) function \(f\) is: \[h'(x) = -\frac{f'(x)}{(f(x))^2},\] wherever \(f\) is nonzero.
In Leibniz's notation, this formula is written: \[\frac{d\left(\frac{1}{f}\right)}{dx} = -\frac{1}{f^2}\frac{df}{dx}.\]
The reciprocal rule can be derived either from the quotient rule or from the combination of power rule and chain rule.
Quotient rule
If \(f\) and \(g\) are functions, then: \[\left(\frac{f}{g}\right)' = \frac{f'g - g'f}{g^2},\] wherever \(g\) is nonzero.
This can be derived from the product rule and the reciprocal rule.
Generalized power rule
The elementary power rule generalizes considerably. The most general power rule is the functional power rule: for any functions \(f\) and \(g\), \[(f^g)' = \left(e^{g\ln f}\right)' = f^g\left(f'{g \over f} + g'\ln f\right),\quad\] wherever both sides are well defined.
Special cases:
- If \(f(x)=x^a\), then \(f'(x)=ax^{a-1}\) when \(a\) is any nonzero real number and \(x\) is positive.
- The reciprocal rule may be derived as the special case where \(g(x)=-1\!\).
Derivatives of exponential and logarithmic functions
\[\frac{d}{dx}\left(c^{ax}\right) = {ac^{ax} \ln c } ,\qquad c > 0.\] The equation above is true for all \(c\), but the derivative for \(c<0\) yields a complex number.
\[\frac{d}{dx}\left(e^{ax}\right) = ae^{ax}.\]
\[\frac{d}{dx}\left( \log_c x\right) = {1 \over x \ln c} , \qquad c > 1.\] The equation above is also true for all \(c\) but yields a complex number if \(c<0\).
\[\frac{d}{dx}\left( \ln x\right) = {1 \over x} ,\qquad x > 0.\]
\[\frac{d}{dx}\left( \ln |x|\right) = {1 \over x} ,\qquad x \neq 0.\]
\[\frac{d}{dx}\left( W(x)\right) = {1 \over {x+e^{W(x)}}} ,\qquad x > -{1 \over e},\] where \(W(x)\) is the Lambert W function.
\[\frac{d}{dx}\left( x^x \right) = x^x(1+\ln x).\]
Condensed: the full section is in Wikipedia.
Logarithmic derivatives
The logarithmic derivative is another way of stating the rule for differentiating the logarithm of a function (using the chain rule): \[(\ln f)'= \frac{f'}{f},\] wherever \(f\) is positive.
Logarithmic differentiation is a technique which uses logarithms and its differentiation rules to simplify certain expressions before actually applying the derivative.
Logarithms can be used to remove exponents, convert products into sums, and convert division into subtraction, each of which may lead to a simplified expression for taking derivatives.
Derivatives of trigonometric functions
The derivatives in the table above are for when the range of the inverse secant is \([0,\pi]\) and when the range of the inverse cosecant is \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\).
It is common to additionally define an inverse tangent function with two arguments, \(\arctan(y,x)\). Its value lies in the range \([-\pi,\pi]\) and reflects the quadrant of the point \((x,y)\). For the first and fourth quadrant (i.e., \(x > 0\)), one has \(\arctan(y, x>0) = \arctan(\frac{y}{x})\). Its partial derivatives are: \[\frac{\partial \arctan(y,x)}{\partial y} = \frac{x}{x^2 + y^2} \qquad\text{and}\qquad \frac{\partial \arctan(y,x)}{\partial x} = \frac{-y}{x^2 + y^2}.\]
Gamma function
\[\Gamma(x) = \int_0^\infty t^{x-1} e^{-t}\, dt\] \[\begin{aligned} \Gamma'(x) & = \int_0^\infty t^{x-1} e^{-t} \ln t\,dt \\ & = \Gamma(x) \left(\sum_{n=1}^\infty \left(\ln\left(1 + \dfrac{1}{n}\right) - \dfrac{1}{x + n}\right) - \dfrac{1}{x}\right) \\ & = \Gamma(x) \psi(x), \end{aligned}\] with \(\psi(x)\) being the digamma function, expressed by the parenthesized expression to the right of \(\Gamma(x)\) in the line above.
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Find the derivative of \(f(x)=8.\)
发送答案
This is just a one-step application of the rule:
\[{f}^{'}(x)=0.\] -
Find the derivative of \(g(x)=-3.\)
发送答案
0
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Find \(\frac{d}{dx}({x}^{3}).\)
发送答案
\[\begin{array}{lllll}\frac{d}{dx}({x}^{3}) & =\underset{h\to 0}{\text{lim}}\frac{{(x+h)}^{3}-{x}^{3}}{h} & & & \\ & =\underset{h\to 0}{\text{lim}}\frac{{x}^{3}+3{x}^{2}h+3x{h}^{2}+{h}^{3}-{x}^{3}}{h} & & & \begin{array}{l}\text{Notice that the first term in the expansion of} \\ {(x+h)}^{3}\ \text{is}\ {x}^{3}\ \text{and the second term is}\ 3{x}^{2}h.\ \text{All} \\ \text{other terms contain powers of}\ h\ \text{that are two or} \\ \text{greater.}\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{3{x}^{2}h+3x{h}^{2}+{h}^{3}}{h} & & & \begin{array}{l}\text{In this step the}\ {x}^{3}\ \text{terms have been cancelled,} \\ \text{leaving only terms containing}\ h.\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{h(3{x}^{2}+3xh+{h}^{2})}{h} & & & \text{Factor out the common factor of}\ h. \\ & =\underset{h\to 0}{\text{lim}}(3{x}^{2}+3xh+{h}^{2}) & & & \begin{array}{l}\text{After cancelling the common factor of}\ h,\text{the} \\ \text{only term not containing}\ h\ \text{is}\ 3{x}^{2}.\end{array} \\ & =3{x}^{2} & & & \text{Let}\ h\ \text{go to 0.}\end{array}\] -
Find \(\frac{d}{dx}({x}^{4}).\)
发送答案
\(4{x}^{3}\)
-
Find the derivative of the function \(f(x)={x}^{10}\) by applying the power rule.
发送答案
Using the power rule with \(n=10,\) we obtain
\[f'(x)=10{x}^{10-1}=10{x}^{9}.\] -
Find the derivative of \(f(x)={x}^{7}.\)
发送答案
\({f}^{'}(x)=7{x}^{6}\)
-
Find the derivative of \(g(x)=3{x}^{2}\) and compare it to the derivative of \(f(x)={x}^{2}.\)
发送答案
We use the power rule directly:
\[{g}^{'}(x)=\frac{d}{dx}(3{x}^{2})=3\frac{d}{dx}({x}^{2})=3(2x)=6x.\]Since \(f(x)={x}^{2}\) has derivative \({f}^{'}(x)=2x,\) we see that the derivative of \(g(x)\) is 3 times the derivative of \(f(x).\) This relationship is illustrated in .
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Find the derivative of \(f(x)=2{x}^{5}+7.\)
发送答案
We begin by applying the rule for differentiating the sum of two functions, followed by the rules for differentiating constant multiples of functions and the rule for differentiating powers. To better understand the sequence in which the differentiation rules are applied, we use Leibniz notation throughout the solution:
\[\begin{array}{lllll}{f}^{'}(x) & =\frac{d}{dx}(2{x}^{5}+7) & & & \\ & =\frac{d}{dx}(2{x}^{5})+\frac{d}{dx}(7) & & & \text{Apply the sum rule.} \\ & =2\frac{d}{dx}({x}^{5})+\frac{d}{dx}(7) & & & \text{Apply the constant multiple rule.} \\ & =2(5{x}^{4})+0 & & & \text{Apply the power rule and the constant rule.} \\ & =10{x}^{4}. & & & \text{Simplify.}\end{array}\] -
Find the derivative of \(f(x)=2{x}^{3}-6{x}^{2}+3.\)
发送答案
\({f}^{'}(x)=6{x}^{2}-12x.\)
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Find an equation of the line tangent to the graph of \(f(x)={x}^{2}-4x+6\) at \(x=1.\)
发送答案
To find an equation of the tangent line, we need a point and a slope. To find the point, compute
\[f(1)={1}^{2}-4(1)+6=3.\]This gives us the point \((1,3).\) Since the slope of the tangent line at 1 is \({f}^{'}(1),\) we must first find \({f}^{'}(x).\) Using the definition of a derivative, we have
\[{f}^{'}(x)=2x-4\]so the slope of the tangent line is \({f}^{'}(1)=-2.\) Using the point-slope formula, we see that the equation of the tangent line is
\[y-3=-2(x-1).\]Putting the equation of the line in slope-intercept form, we obtain
\[y=-2x+5.\] -
Find an equation of the line tangent to the graph of \(f(x)=3{x}^{2}-11\) at \(x=2.\) Use the point-slope form.
发送答案
\(y=12x-23\)
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For \(j(x)=f(x)g(x),\) use the product rule to find \({j}^{'}(2)\) if \(f(2)=3,{f}^{'}(2)=-4,g(2)=1,\) and \({g}^{'}(2)=6.\)
发送答案
Since \(j(x)=f(x)g(x),{j}^{'}(x)={f}^{'}(x)g(x)+{g}^{'}(x)f(x),\) and hence
\[{j}^{'}(2)={f}^{'}(2)g(2)+{g}^{'}(2)f(2)=(-4)(1)+(6)(3)=14.\] -
For \(j(x)=({x}^{2}+2)(3{x}^{3}-5x),\) find \({j}^{'}(x)\) by applying the product rule. Check the result by first finding the product and then differentiating.
发送答案
If we set \(f(x)={x}^{2}+2\) and \(g(x)=3{x}^{3}-5x,\) then \({f}^{'}(x)=2x\) and \({g}^{'}(x)=9{x}^{2}-5.\) Thus,
\[{j}^{'}(x)={f}^{'}(x)g(x)+{g}^{'}(x)f(x)=(2x)(3{x}^{3}-5x)+(9{x}^{2}-5)({x}^{2}+2).\]Simplifying, we have
\[{j}^{'}(x)=15{x}^{4}+3{x}^{2}-10.\]To check, we see that \(j(x)=3{x}^{5}+{x}^{3}-10x\) and, consequently, \({j}^{'}(x)=15{x}^{4}+3{x}^{2}-10.\)
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Use the product rule to obtain the derivative of \(j(x)=2{x}^{5}(4{x}^{2}+x).\)
发送答案
\({j}^{'}(x)=10{x}^{4}(4{x}^{2}+x)+(8x+1)(2{x}^{5})=56{x}^{6}+12{x}^{5}.\)
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Use the quotient rule to find the derivative of \(k(x)=\frac{5{x}^{2}}{4x+3}.\)
发送答案
Let \(f(x)=5{x}^{2}\) and \(g(x)=4x+3.\) Thus, \({f}^{'}(x)=10x\) and \({g}^{'}(x)=4.\) Substituting into the quotient rule, we have
\[{k}^{'}(x)=\frac{{f}^{'}(x)g(x)-{g}^{'}(x)f(x)}{{(g(x))}^{2}}=\frac{10x(4x+3)-4(5{x}^{2})}{{(4x+3)}^{2}}.\]Simplifying, we obtain
\[{k}^{'}(x)=\frac{20{x}^{2}+30x}{{(4x+3)}^{2}}.\] -
Find the derivative of \(h(x)=\frac{3x+1}{4x-3}.\)
发送答案
\({k}^{'}(x)=-\frac{13}{{(4x-3)}^{2}}.\)
-
Find \(\frac{d}{dx}({x}^{-4}).\)
发送答案
By applying the extended power rule with \(k=-4,\) we obtain
\[\frac{d}{dx}({x}^{-4})=-4{x}^{-4-1}=-4{x}^{-5}.\] -
Use the extended power rule and the constant multiple rule to find the derivative of \(f(x)=\frac{6}{{x}^{2}}.\)
发送答案
It may seem tempting to use the quotient rule to find this derivative, and it would certainly not be incorrect to do so. However, it is far easier to differentiate this function by first rewriting it as \(f(x)=6{x}^{-2}.\)
\[\begin{array}{lllll}{f}^{'}(x) & =\frac{d}{dx}(\frac{6}{{x}^{2}})=\frac{d}{dx}(6{x}^{-2}) & & & \text{Rewrite}\ \frac{6}{{x}^{2}}\ \text{as}\ 6{x}^{-2}. \\ & =6\frac{d}{dx}({x}^{-2}) & & & \text{Apply the constant multiple rule.} \\ & =6(-2{x}^{-3}) & & & \text{Use the extended power rule to differentiate}\ {x}^{-2}. \\ & =-12{x}^{-3} & & & \text{Simplify.}\end{array}\] -
Find the derivative of \(g(x)=\frac{1}{{x}^{7}}\) using the extended power rule.
发送答案
\({g}^{'}(x)=-7{x}^{-8}.\)
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For \(k(x)=3h(x)+{x}^{2}g(x),\) find \({k}^{'}(x).\)
发送答案
Finding this derivative requires the sum rule, the constant multiple rule, and the product rule.
\[\begin{array}{lllll}{k}^{'}(x) & =\frac{d}{dx}(3h(x)+{x}^{2}g(x))=\frac{d}{dx}(3h(x))+\frac{d}{dx}({x}^{2}g(x)) & & & \text{Apply the sum rule.} \\ & =3\frac{d}{dx}(h(x))+(\frac{d}{dx}({x}^{2})g(x)+\frac{d}{dx}(g(x)){x}^{2}) & & & \begin{array}{l}\text{Apply the constant multiple rule to} \\ \text{differentiate}\ 3h(x)\ \text{and the product} \\ \text{rule to differentiate}\ {x}^{2}g(x).\end{array} \\ & =3{h}^{'}(x)+2xg(x)+{x}^{2}{g}^{'}(x) & & & \end{array}\] -
For \(k(x)=f(x)g(x)h(x),\) express \({k}^{'}(x)\) in terms of \(f(x),g(x),h(x),\) and their derivatives.
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We can think of the function \(k(x)\) as the product of the function \(f(x)g(x)\) and the function \(h(x).\) That is, \(k(x)=(f(x)g(x))\cdot h(x).\) Thus,
\[\begin{array}{lllll}{k}^{'}(x) & =\frac{d}{dx}(f(x)g(x))\cdot h(x)+\frac{d}{dx}(h(x))\cdot (f(x)g(x)) & & & \begin{array}{l}\text{Apply the product rule to the product} \\ \text{of}\ f(x)g(x)\ \text{and}\ h(x).\end{array} \\ & =({f}^{'}(x)g(x)+{g}^{'}(x)f(x))h(x)+{h}^{'}(x)f(x)g(x) & & & \text{Apply the product rule to}\ f(x)g(x). \\ & ={f}^{'}(x)g(x)h(x)+f(x){g}^{'}(x)h(x)+f(x)g(x){h}^{'}(x). & & & \text{Simplify.}\end{array}\] -
For \(h(x)=\frac{2{x}^{3}k(x)}{3x+2},\) find \({h}^{'}(x).\)
发送答案
This procedure is typical for finding the derivative of a rational function.
\[\begin{array}{llllll}{h}^{'}(x) & =\frac{\frac{d}{dx}(2{x}^{3}k(x))\cdot (3x+2)-\frac{d}{dx}(3x+2)\cdot (2{x}^{3}k(x))}{{(3x+2)}^{2}} & & & \text{Apply the quotient rule.} & \ \\ & =\frac{(6{x}^{2}k(x)+{k}^{'}(x)\cdot 2{x}^{3})(3x+2)-3(2{x}^{3}k(x))}{{(3x+2)}^{2}} & & & \begin{array}{l}\text{Apply the product rule to find} \\ \frac{d}{dx}(2{x}^{3}k(x)).\ \text{Use}\ \frac{d}{dx}(3x+2)=3.\end{array} & \ \\ & =\frac{-6{x}^{3}k(x)+18{x}^{3}k(x)+12{x}^{2}k(x)+6{x}^{4}{k}^{'}(x)+4{x}^{3}{k}^{'}(x)}{{(3x+2)}^{2}} & & & \text{Simplify.} & \ \\ & =\frac{12k\left(x\right)\left({x}^{3}+{x}^{2}\right)+2k'\left(x\right)\left(3{x}^{4}+2{x}^{3}\right)}{{\left(3x+2\right)}^{2}} & \ & \ & \ & \ \end{array}\] -
Find \(\frac{d}{dx}(3f(x)-2g(x)).\)
发送答案
\(3{f}^{'}(x)-2{g}^{'}(x).\)
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Determine the values of \(x\) for which \(f(x)={x}^{3}-7{x}^{2}+8x+1\) has a horizontal tangent line.
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To find the values of \(x\) for which \(f(x)\) has a horizontal tangent line, we must solve \({f}^{'}(x)=0.\) Since
\[{f}^{'}(x)=3{x}^{2}-14x+8=(3x-2)(x-4),\]we must solve \((3x-2)(x-4)=0.\) Thus we see that the function has horizontal tangent lines at \(x=\frac{2}{3}\) and \(x=4\) as shown in the following graph.
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The position of an object on a coordinate axis at time \(t\) is given by \(s(t)=\frac{t}{{t}^{2}+1}.\) What is the initial velocity of the object?
发送答案
Since the initial velocity is \(v(0)={s}^{'}(0),\) begin by finding \({s}^{'}(t)\) by applying the quotient rule:
\[{s}^{'}(t)=\frac{1({t}^{2}+1)-2t(t)}{{({t}^{2}+1)}^{2}}=\frac{1-{t}^{2}}{{({t}^{2}+1)}^{2}}.\]After evaluating, we see that \(v(0)=1.\)
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Find the values of \(x\) for which the graph of \(f(x)=4{x}^{2}-3x+2\) has a tangent line parallel to the line \(y=2x+3.\)
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\(\frac{5}{8}\)
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\(f(x)={x}^{7}+10\)
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\(f(x)=5{x}^{3}-x+1\)
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\({f}^{'}(x)=15{x}^{2}-1\)
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\(f(x)=4{x}^{2}-7x\)
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\(f(x)=8{x}^{4}+9{x}^{2}-1\)
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\({f}^{'}(x)=32{x}^{3}+18x\)
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\(f(x)={x}^{4}+\frac{2}{x}\)
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\(f(x)=3x(18{x}^{4}+\frac{13}{x+1})\)
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\({f}^{'}(x)=270{x}^{4}+\frac{39}{{(x+1)}^{2}}\)
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\(f(x)=(x+2)(2{x}^{2}-3)\)
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\(f(x)={x}^{2}(\frac{2}{{x}^{2}}+\frac{5}{{x}^{3}})\)
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\({f}^{'}(x)=\frac{-5}{{x}^{2}}\)
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\(f(x)=\frac{{x}^{3}+2{x}^{2}-4}{3}\)
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\(f(x)=\frac{4{x}^{3}-2x+1}{{x}^{2}}\)
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\({f}^{'}(x)=\frac{4{x}^{4}+2{x}^{2}-2x}{{x}^{4}}\)
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\(f(x)=\frac{{x}^{2}+4}{{x}^{2}-4}\)
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\(f(x)=\frac{x+9}{{x}^{2}-7x+1}\)
发送答案
\({f}^{'}(x)=\frac{\text{-}{x}^{2}-18x+64}{{({x}^{2}-7x+1)}^{2}}\)
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[T] \(y=3{x}^{2}+4x+1\) at \((0,1)\)
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[T] \(y=\frac{2}{{x}^{2}}+1\) at \((1,3)\)
发送答案
\(T(x)=-4x+7\)
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人们对人们提出的问题
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.
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