| The Chain Rule Basics | |
| The Equation of the Tangent Line with the Chain Rule | |
| More Practice |
The chain rule says when we’re taking the derivative, if there’s something other than $ \boldsymbol {x}$, like in the parenthesis or under a radical sign, for example, we have to multiply what we get by the derivative of what’s inside the parentheses. It all has to do with Composite Functions, since $ \displaystyle \frac{{dy}}{{dx}}=\frac{{dy}}{{du}}\cdot \frac{{du}}{{dx}}$. Note that we’ll learn how to “undo” the chain rule here in the U-Substitution Integration section.
Think of it this way when we’re thinking of rates of change, or derivatives: if we are running twice as fast as Person A, and then Person B is running three times as fast as us, Person B is running six times as fast as Person A. It’s all about relativity! Here is what it looks like in Theorem form:
If $ \displaystyle y=f\left( u \right)$ and $ u=g\left( x \right)$ are differentiable and $ y=f\left( {g\left( x \right)} \right)$, then:
$ \displaystyle \frac{{dy}}{{dx}}=\frac{{dy}}{{du}}\cdot \frac{{du}}{{dx}}$, or
$ \displaystyle \frac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right]={f}’\left( {g\left( x \right)} \right){g}’\left( x \right)$ (more simplified): $ \displaystyle \frac{d}{{dx}}\left[ {f\left( u \right)} \right]={f}’\left( u \right){u}’$
We’ve actually been using the chain rule all along, since the derivative of an expression with just an $ \boldsymbol {x}$ in it is just 1, so we are multiplying by 1. For example, if $ \displaystyle y={{x}^{2}},\,\,\,\,\,{y}’=2x\cdot \frac{{d\left( x \right)}}{{dx}}=2x\cdot 1=2x$.
Do you see how when we take the derivative of the “outside function” and there’s something other than just $ \boldsymbol {x}$ in the argument (for example, in parentheses, under a radical sign, or in a trigonometric function), we have to take the derivative again of this “inside function”? In a nutshell, we are taking the derivative of the “outside function” and multiplying this by the derivative of the “inside” function(s). That’s pretty much it!
In the problems below, see how we take the derivative again of what’s in red? And sometimes, again, what’s in blue? Yes, sometimes we have to use the chain rule twice, in the cases where we have a function inside a function inside another function. We could theoretically take the chain rule a very large number of times, with one derivative!
| Function | Derivative Using Chain Rule | Notes |
| $ \displaystyle f\left( x \right)={{\left( {5x-1} \right)}^{8}}$ | $ \displaystyle \begin{align}{f}’\left( x \right)&=8{{\left( {\color{red}{{5x-1}}} \right)}^{7}}\cdot \color{red}{5}\\&=40{{\left( {5x-1} \right)}^{7}}\end{align}$ | Since the $ \left( {5x-1} \right)$ is the inner function, after using the Power Rule, multiply by the derivative of that function, which is 5. |
| $ \displaystyle f\left( x \right)={{\left( {{{x}^{4}}-1} \right)}^{3}}$ | $ \displaystyle \begin{align}{f}’\left( x \right)&=3{{\left( {\color{red}{{{{x}^{4}}-1}}} \right)}^{2}}\cdot \left( {\color{red}{{4{{x}^{3}}}}} \right)\\&=12{{x}^{3}}{{\left( {{{x}^{4}}-1} \right)}^{2}}\end{align}$ | Since the $ \left( {{{x}^{4}}-1} \right)$ is the inner function, after using the Power Rule, multiply by the derivative of that function, which is $ 4{{x}^{3}}$. |
| $ \displaystyle \begin{array}{l}g\left( x \right)=\sqrt[4]{{16-{{x}^{3}}}}\\g\left( x \right)={{\left( {16-{{x}^{3}}} \right)}^{{\frac{1}{4}}}}\end{array}$ | $ \displaystyle \begin{align}{l}{g}’\left( x \right)&=\frac{1}{4}{{\left( {\color{red}{{16-{{x}^{3}}}}} \right)}^{{-\frac{3}{4}}}}\cdot \left( {\color{red}{{-3{{x}^{2}}}}} \right)\\&=-\frac{{3{{x}^{2}}}}{{4{{{\left( {16-{{x}^{3}}} \right)}}^{{\frac{3}{4}}}}}}=-\frac{{3{{x}^{2}}}}{{4\,\sqrt[4]{{{{{\left( {16-{{x}^{3}}} \right)}}^{3}}}}}}\end{align}$ | Since the $ \left( {16-{{x}^{3}}} \right)$ is the inner function, after using the Power Rule, multiply by the derivative of that function, which is $ -3{{x}^{2}}$. |
| $ \displaystyle \begin{array}{l}f\left( t \right)={{\left( {3t+4} \right)}^{4}}\sqrt{{3t-2}}\\f\left( t \right)={{\left( {3t+4} \right)}^{4}}{{\left( {3t-2} \right)}^{{\frac{1}{2}}}}\end{array}$ | $ \displaystyle \begin{align}{f}’\left( t \right)&={{\left( {3t+4} \right)}^{4}}\left( {\frac{1}{2}} \right){{\left( {\color{red}{{3t-2}}} \right)}^{{-\frac{1}{2}}}}\cdot \left( {\color{red}{3}} \right)\\&\,\,\,\,\,\,\,+{{\left( {3t-2} \right)}^{{\frac{1}{2}}}}\cdot 4{{\left( {\color{red}{{3t+4}}} \right)}^{3}}\cdot \left( {\color{red}{3}} \right)\\&=\frac{3}{2}{{\left( {3t+4} \right)}^{4}}{{\left( {3t-2} \right)}^{{-\frac{1}{2}}}}+12{{\left( {3t-2} \right)}^{{\frac{1}{2}}}}{{\left( {3t+4} \right)}^{3}}\\&=\frac{3}{2}{{\left( {3t+4} \right)}^{3}}{{\left( {3t-2} \right)}^{{-\frac{1}{2}}}}\left( {\left( {3t+4} \right)+8\left( {3t-2} \right)} \right)\\&=\frac{3}{2}{{\left( {3t+4} \right)}^{3}}{{\left( {3t-2} \right)}^{{-\frac{1}{2}}}}\left( {27t-12} \right)\\&=\frac{{3{{{\left( {3t+4} \right)}}^{3}}\left( {27t-12} \right)}}{{2\sqrt{{3t-2}}}}=\frac{{9{{{\left( {3t+4} \right)}}^{3}}\left( {9t-4} \right)}}{{2\sqrt{{3t-2}}}}\end{align}$ | Use the Product Rule, since there are $ t$’s in both expressions. Since $ \left( {3t+4} \right)$ and $ \left( {3t-2} \right)$ are the inner functions, multiply each by their derivative.
Note that we also took out the Greatest Common Factor (GCF) $ \displaystyle \frac{3}{2}{{\left( {3t+4} \right)}^{3}}{{\left( {3t-2} \right)}^{{-\frac{1}{2}}}}$, so we could simplify the expression. (Remember, with the GCF, take out factors with the smallest exponent.) The reason we also took out a $ \displaystyle \frac{3}{2}$ is because it’s the GCF of $ \displaystyle \frac{3}{2}$ and $ \displaystyle \frac{{24}}{2}\,\,(12)$. |
| $ \displaystyle y=\cos \left( {4x} \right)$ | $ \displaystyle \begin{array}{l}{y}’=-\sin \left( {\color{red}{{4x}}} \right)\cdot \color{red}{4}\\=-4\sin \left( {4x} \right)\end{array}$ | Since the $ \left( {4x} \right)$ is the inner function (the argument of $ \text{sin}$), multiply by the derivative of that function, which is 4. |
| $ \displaystyle g\left( x \right)=\cos \left( {\tan x} \right)$ | $ \displaystyle \begin{align}{g}’\left( x \right)&=-\sin \left( {\color{red}{{\tan x}}} \right)\cdot \color{red}{{{{{\sec }}^{2}}x}}\\&=-{{\sec }^{2}}x\cdot \sin \left( {\tan x} \right)\end{align}$ | Since the $ \left( {\tan x} \right)$ is the inner function (the argument of $ \text{cos}$), multiply by the derivative of that function, which is $ \displaystyle {{\sec }^{2}}x$. |
| $ \displaystyle \begin{array}{l}f\left( x \right)={{\sec }^{3}}\left( {\pi x} \right)\\f\left( x \right)={{\left[ {\sec \left( {\pi x} \right)} \right]}^{3}}\end{array}$ (Note second way to write this) | $ \displaystyle \begin{align}{f}’\left( x \right)&=3\,{{\color{red}{{\sec }}}^{2}}\left( {\color{blue}{{\pi x}}} \right)\cdot \left( {\color{red}{{\sec \left( {\color{blue}{{\pi x}}} \right)\tan \left( {\color{blue}{{\pi x}}} \right)}}} \right)\color{blue}{\pi }\\&=3\pi {{\sec }^{3}}\left( {\pi x} \right)\tan \left( {\pi x} \right)\end{align}$ | This one’s a little tricky, since we have to use the Chain Rule twice. To take the derivative of $ {{\left[ {\sec \left( {\pi x} \right)} \right]}^{3}}$, multiply by the derivative of $ \sec \left( {\pi x} \right)$, which is $ \sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi $ (again, multiply by the derivative of $ \pi x$, which is $ \pi $). |
| $ \displaystyle \begin{array}{l}f\left( \theta \right)=2{{\cot }^{2}}\left( {2\theta } \right)+\theta \\f\left( \theta \right)=2{{\left[ {\cot \left( {2\theta } \right)} \right]}^{2}}+\theta \end{array}$ | $ \displaystyle \begin{align}{f}’\left( \theta \right)=&4\,\color{red}{{\cot }}\left( {\color{blue}{{2\theta }}} \right)\cdot \color{red}{{-{{{\csc }}^{2}}\left( {\color{blue}{{2\theta }}} \right)}}\cdot \color{blue}{2}+1\\&=1-8{{\csc }^{2}}\left( {2\theta } \right)\cot \left( {2\theta } \right)\end{align}$ | This is another one where we have to use the Chain Rule twice. |
Here’s one more problem, where we have to think about how the chain rule works:
| Chain Rule Problem | ||||||||||||||||||||||||||
| The graph and table for functions $ f$ and $ g$ are below. Let $ p\left( x \right)=f\left( {g\left( x \right)} \right)$ and $ q\left( x \right)=g\left( {f\left( x \right)} \right)$. Find $ {p}’\left( 4 \right)\text{ and }{q}’\left( {-1} \right)$, given these derivatives exist. | ||||||||||||||||||||||||||
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| |||||||||||||||||||||||||
| $ \displaystyle \begin{array}{c}\color{#800000}{{\text{Find}\,\,{p}’\left( 4 \right):}}\\p\left( x \right)=f\left( {g\left( x \right)} \right)\\{p}’\left( x \right)={f}’\left( {g\left( x \right)} \right)\cdot {g}’\left( x \right)\\{p}’\left( 4 \right)={f}’\left( {g\left( 4 \right)} \right)\cdot {g}’\left( 4 \right)\\{p}’\left( 4 \right)={f}’\left( 6 \right)\cdot {g}’\left( 4 \right)\\{p}’\left( 4 \right)=0\cdot 3=0\end{array}$ $ \displaystyle \begin{array}{c}\color{#800000}{{\text{Find}\,\,{q}’\left( {-1} \right):}}\\q\left( x \right)=g\left( {f\left( x \right)} \right)\\{q}’\left( x \right)={g}’\left( {f\left( x \right)} \right)\cdot {f}’\left( x \right)\\{q}’\left( {-1} \right)={g}’\left( {f\left( {-1} \right)} \right)\cdot {f}’\left( {-1} \right)\\{q}’\left( {-1} \right)={g}’\left( 2 \right)\cdot {f}’\left( {-1} \right)\\{g}’\left( 2 \right)\,\,\text{doesn }\!\!’\!\!\text{ t exist}\,\,(\text{sharp turn)}\\\text{Therefore, }{q}’\left( {-1} \right)\,\,\text{doesn }\!\!’\!\!\text{ t exist}\end{array}$ | ||||||||||||||||||||||||||
The Equation of the Tangent Line with the Chain Rule
Here are a few problems where we use the chain rule to find an equation of the tangent line to the graph $ f$ at the given point. Note that we saw more of these problems here in the Equation of the Tangent Line, Tangent Line Approximation, and Rates of Change Section.
| Function | Equation of Tangent Line |
| Find the equation of the tangent line to the graph of $ f$ at the given $ x$ value:
$ \begin{array}{c}f\left( x \right)={{\left( {5{{x}^{4}}-2} \right)}^{3}}\\x=1\end{array}$ | $ \displaystyle {f}’\left( x \right)=3{{\left( {5{{x}^{4}}-2} \right)}^{2}}\left( {20{{x}^{3}}} \right)=60{{x}^{3}}{{\left( {5{{x}^{4}}-2} \right)}^{2}}$
The slope of the function is $ \displaystyle 60{{x}^{3}}{{\left( {5{{x}^{4}}-2} \right)}^{2}}$, and at $ x=1$, $ \displaystyle y={{\left( {5{{{\left( 1 \right)}}^{4}}-2} \right)}^{3}}=27$. At point $ \left( {1,27} \right)$, the slope is $ \displaystyle 60{{\left( 1 \right)}^{3}}{{\left[ {5{{{\left( 1 \right)}}^{4}}-2} \right]}^{2}}=540$. Use either the slope-intercept or point-slope method to find the equation of the line (let’s use slope-intercept): $ y=mx+b;\,\,y=540x+b$. Plug in point $ \left( {1,27} \right)$ and solve for $ b$: $ 27=540\left( 1 \right)+b;\,\,\,b=-513$. The equation of the tangent line to $ f\left( x \right)={{\left( {5{{x}^{4}}-2} \right)}^{3}}$ at $ x=1$ is $ \,y=540x-513$. |
| Find the equation of the tangent line to the graph of $ f$ at the given point:
$ f\left( \theta \right)=\cos \left( {5\theta } \right)$ $ \displaystyle \left( {\frac{\pi }{2},0} \right)$ | $ \displaystyle {f}’\left( x \right)=-5\sin \left( {5\theta } \right)$
The slope of the function is $ \displaystyle -5\sin \left( {5\theta } \right)$, so at point $ \displaystyle \left( {\frac{\pi }{2},0} \right)$, the slope is $ \displaystyle -5\sin \left( {5\cdot \frac{\pi }{2}} \right)=-5$. Use either the slope-intercept or point-slope method to find the equation of the line (let’s use point-slope): $ \displaystyle y-0=-5\left( {x-\frac{\pi }{2}} \right);\,\,y=-5x+\frac{{5\pi }}{2}$. The equation of the tangent line to $ f\left( \theta \right)=\cos \left( {5\theta } \right)$ at the point $ \displaystyle \left( {\frac{\pi }{2},0} \right)$ is $ \displaystyle y=-5x+\frac{{5\pi }}{2}$. |
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