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The Chain Rule

April 9th, 2009 |

Author: Merrin

The chain rule is undoubtedly the most difficult differentiation rule to

learn, but you need it to differentiate more complicated functions that result from compositions. You simply can’t solve the harder problems without it. The chain rule can be stated as

$\cfrac{d}{dx}f(u(x))=\cfrac{df}{du}\cfrac{du}{dx}$

We can find the derivatives for a few examples using the

chain rule and regular methods as a comparison to verify the rule.

Example 1. Find the derivative of f(x)=sin(2x).

Solution 1. We identify u(x)=2x and f(u)=sin u. Applying the chain

rule we have

$\cfrac{d}{dx}f(u(x))=\cfrac{df}{du}\cfrac{du}{dx}=(\cos u )(2)=2\cos (2x)$

Alternatively we can use our trigonometric identities to simply the derivative.

$f(x)=\sin (2x)=2\sin x\cos x$

With this form we can use the product rule.

$\cfrac{df}{dx}=2(\sin x)'(\cos x)+2(\sin x)(\cos x)'=2\cos^2x-2\sin^2x=2\cos(2x)$

Of course we get the same answer.

Example 2. Another example which can be done several ways is to find the

derivative of

$f(x)=(1+x^2)^2$

Solution 2. One method is to multiply everything out and then take the derivative of the polynomial. We should already know how to carry out this method by now.

$f(x)=1+2x^2+x^4$

$f'(x)=4x+4x^3=4x(1+x^2)$

This problem can also be solved with the chain rule. Using u(x)=1+x^2 we

can express f(x) as f(u)=u^2

$\cfrac{df}{du}\cfrac{du}{dx}=2u(2x)=4x(1+x^2 )$

The chain rule has been verified for several example. Now is a good time

to prove the chain rule.

Proof 1. To prove the chain rule we apply the definition of the derivative and multiply and divide by the same quantity, dg.

${y}= {f(u(x))} \\ {\cfrac{dy}{dx} }={\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(u(x+\Delta x)-f(u(x))}{\Delta x} } \\ = {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(u(x+\Delta x)-f(u(x))}{u(x+\Delta x)-u(x)} \cfrac{u(x+\Delta x)-u(x)}{\Delta x} } \\ = {\mathop{\lim }\limits_{\Delta x\to 0} \cfrac{f(u(x+\Delta x))-f(u(x))}{u(x+\Delta x)-u(x)} \cdot \mathop{\lim }\limits_{\Delta x\to 0} \cfrac{u(x+\Delta x)-u(x)}{\Delta x} } \\ ={\mathop{\lim }\limits_{\Delta g\to 0} \cfrac{f(u(x)+\Delta u)-f(u(x))}{\Delta u} \cdot \cfrac{du}{dx} } \\ = {\cfrac{df}{du} \cfrac{du}{dx} }$

Half the problem of using the chain rule is deciding what function to pick for u(x) within the whole function. A good choice generally will leave the main function as one of the elementary functions we already know how to differentiate.

Example 3. Find the composition of functions which is easily dealt with using the chain rule for each example differentiate the functions using the chain rule.

$\begin{array}{l} {f_{1} (x)=\sin (5x)} \\ {f_{2} (x)=(1+x^{2} )^{5} } \\ {f_{3} (x)=\, \ln (1+x)} \\ {f_{4} (x)=\sin (\cos x)} \\ {f_{5} (x)=\cfrac{1}{\sqrt{1-x^{2} } } } \end{array}$

Solution 3. We notice that the functions are more complicated but we know how to differentiate pieces of the functions individually if we choose the right composition.

$\begin{array}{ll}f_{1} (u)=\sin u \qquad & u(x)=5x \\f_{2} (u)=u^{5} \qquad & u(x)=1+x^{2} \\f_{3} (u)=\ln (u) \qquad & u(x)=1+x \\f_{4} (u)=\sin (u) \qquad & u(x) =\cos x \\f_{5} (u)=u^{-1/2} \qquad & u(x)=1-x^{2}\end{array}$

Notice how we know how to differentiate all of the functions with respect to u(x) and we know how to differentiate all of the u(x) with respect to x. Now we apply the chain rule.

$\begin{array}{ll}f'_{1} (u)=\cos (u) \qquad & u'_{1}(x)=5 \\f'_{2} (u)=5u^{4} \qquad & u'_{2}(x)=2x \\f'_{3} (u)=\, u^{-1} \qquad & u'_{3}(x)=1 \\f'_{4} (u)=\cos (u) \qquad & u'_{4}(x)=-\sin x \\f'_{5} (u)=-\frac{1}{2}u^{-3/2} \qquad & u'_{5}(x)=-2x\end{array}$

Now multiply two two derivatives together to get the total derivative.

$\begin{array}{l}f'_{1} (x)= 5\cos(5x) \\f'_{2} (x)=10x(1+x^{2} )^{4} \\f'_{3} (x)= {1}/({1+x}) \\f'_{4} (x)=-\sin x\cos (\cos x) \\f'_{5} (x)=x(1-x^{2})^{-3/2}\end{array}$

Example 4. What is the chain rule of the composition of three functions?

Solution 4. The chain rule is applicable recursively for compositions beyond two functions thus it is called the “chain” rule

${y} ={f(g(h(x)))=f(g)} \\ {\cfrac{dy}{dx} } ={\cfrac{df}{dg} \cfrac{dg}{dx} =\cfrac{df}{dg} \cdot \left(\cfrac{d}{dx} g(h(x))\right)=\cfrac{df}{dg} \cfrac{dg}{dh} \cfrac{dh}{dx} }$

Example 5. Find the derivative of f(x).

$f(x)=(\alpha+\beta x^{\gamma} )^{\delta}$

Solution 5. We can set the quantities inside the parentheses as one function then we can write.

$f(u)=u^{\delta}$

Normally if we took the derivative of this function with respect to u we could just write.

$\cfrac{df}{du} =\delta u^{\delta-1}$

But we want to find the derivative of f with respect to x so we use the chain rule.

$\cfrac{df}{dx} =\cfrac{df}{du} \cfrac{du}{dx} =\delta u^{\delta-1} \cfrac{d}{dx} (\alpha+\beta x^{\gamma} )=\delta u^{\delta-1} (0+\beta \gamma x^{\gamma-1} )= \beta \gamma \delta x^{\gamma-1} (\alpha +\beta x^{\gamma} )^{\delta-1}$

Example 6. Find the derivative of f(x) where

$f(x)=\ln (\sin (5x)))$

Solution 6. Here, we can write the function three levels of composition

deep, where

$f(u)=\ln u(x)\qquad u(v)=\sin (v)\quad \quad v(x)=5x$

$\cfrac{d}{dx} f(u(v)))=\cfrac{df}{du} \cfrac{du}{dv} \cfrac{dv}{dx} =\cfrac{1}{u} \cos (v)\cdot 5=5\cfrac{\cos (5x)}{\sin (5x)} =5\cot (5x)$

In the previous example, there was an internal function of 5x and we got a factor of 5 multiplying the rest. It is quite common in problems for there to be an extra scale factor multiplying x. Here are some other examples of a constant multiplying x.

Example 7. Multiplying derivatives by a scale factor.

${f(x)=\sin (ax)\quad \, \cfrac{d}{dx} \sin (ax)=a\cos (ax)} \\ {f(x)=e^{ax} \quad \cfrac{d}{dx} e^{ax} =ae^{ax} } \\ {f(x)=\ln (ax)=\ln a+\ln x\quad \cfrac{d}{dx} \ln (ax)=\cfrac{1}{x} }$

What is special about the logarithm, how come the derivative has no factor of a in front of it like the others?

Solution 7. Well the factor of a canceled itself out.

This is a special property of the logarithm.

$\cfrac{d}{dx} \ln (ax)=\cfrac{1}{ax} \cfrac{d}{dx} (ax)=\cfrac{1}{ax} \cdot a=\cfrac{1}{x}$

Another way to see that there is no factor is to break the logarithm up into a constant part and ln x. Since the derivative of a constant is zero there is no factor multiplying the derivative.

Example 8. Find the derivative of

$f(x)=\ln (2+\sin (e^{x} ))$

Solution 8. Here we won’t use any fancy notation, we will just keep

using the variable u until we run out of things to differentiate.There are many composition of functions presented here. We can start to break down $f(x)$ by defining

$f(u)=\ln u\to u=2+\sin (e^{x} )$

Now apply the chain rule with respect to u.

$\cfrac{d}{dx} f(u)=\cfrac{1}{u} \cfrac{d}{dx}(2+\sin (e^{x}))$

$\cfrac{d}{dx}(2+\sin(e^x))=0+\cos(e^x)(e^x)'=e^x (\cos (e^x))$

$\cfrac{df}{dx}=\cfrac{e^x\cos( e^{x})}{2+\sin(e^x)}$

Try to apply the chain rule in your head. This is just a matter of drilling.

The chain rule is useful because we do calculus on bigger and bigger functions the more advanced we get. The chain rule provides a method to break down the process of differentiation into a finite number of pieces that are straightforward. In general, differentiation is a straightforward process. There aren’t volumes of books devoted to tables of derivatives however there are such books on integration which is more of an art which we get to in the next chapter. This is because differentiation has been broken down into a set number of rules that any person or computer can calculate given enough time. Differentiation is solvable.

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