2.4: Differentiation Techniques - The Chain Rule
- Page ID
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- This is a standard Chain Rule section. As with everything up this point, derive theorems and tie in old Algebra and Trig as often as possible.
The following is a list of learning objectives for this section.
- Learning Objectives (click to expand)
-
- State the Chain Rule using both Lagrange and Leibniz notations.
- Use the Chain Rule combined with the Power Rule.
- Apply the Chain Rule and the Product/Quotient Rules correctly when necessary.
- Describe the proof of the Chain Rule.
We have seen the techniques for differentiating basic functions (\(x^n,\sin x,\cos x\), etc.) as well as sums, differences, products, quotients, and constant multiples of these functions. However, these techniques do not allow us to differentiate compositions of functions, such as \(h(x)=\sin(x^3)\) or \(k(x)=\sqrt{3x^2+1}\). In this section, we study the rule for finding the derivative of the composition of two or more functions.
Deriving the Chain Rule
When we have a function composed of two or more functions, we can use all the techniques we have already learned to differentiate it. However, using all those techniques to break down a function into simpler parts that we can differentiate can be challenging. Instead, we use the Chain Rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
To put this rule into context, let's take a look at an example: \(h(x)=\sin(x^3)\). We can think of the derivative of this function with respect to \(x\) as the rate of change of \(\sin(x^3)\) relative to the change in \(x\). Consequently, we want to know how \(\sin(x^3)\) changes as \(x\) changes. We can think of this event as a chain reaction: As \(x\) changes, \(x^3\) changes, which leads to a change in \(\sin(x^3)\). This chain reaction hints at what is involved in computing the derivative of \(\sin(x^3)\). First of all, a change in \(x\) forcing a change in \(x^3\) suggests that somehow the derivative of \(x^3\) is involved. In addition, the change in \(x^3\) forcing a change in \(\sin(x^3)\) suggests that the derivative of \(\sin(u)\) with respect to \(u\), where \(u=x^3\), is also part of the final derivative.
We can take a more formal look at the derivative of \(h(x)=\sin(x^3)\) by setting up the limit that would give us the derivative at a specific value \(a\) in the domain of \(h(x)=\sin(x^3)\).\[h^{\prime}(a)=\lim_{x \to a}\dfrac{\sin(x^3)-\sin(a^3)}{x-a}\nonumber\]This expression does not seem particularly helpful; however, we can modify it by multiplying and dividing by the expression \(x^3-a^3\) to obtain\[h^{\prime}(a)=\lim_{x \to a}\dfrac{\sin(x^3)-\sin(a^3)}{x^3-a^3} \cdot \dfrac{x^3-a^3}{x-a}.\nonumber\]From the definition of the derivative, we can see that the second factor is the derivative of \(x^3\) at \(x=a\). That is,\[\lim_{x \to a}\dfrac{x^3-a^3}{x-a}=\left. \dfrac{d}{dx}(x^3)\right|_{x=a}=3a^2.\nonumber\]However, recognizing that the first term is also a derivative might be a little more challenging. We can see this by letting \(u=x^3\) and observing that as \(x \to a,u \to a^3\):\[\begin{array}{rcl} \displaystyle \lim_{x \to a}\dfrac{\sin(x^3)-\sin(a^3)}{x^3-a^3} & = & \displaystyle \lim_{u \to a^3}\dfrac{\sin u-\sin(a^3)}{u-a^3} \\[16pt] & = & \left. \dfrac{d}{du}(\sin u)\right|_{u=a^3} \\[16pt] & = & \cos(a^3) \\[16pt] \end{array}\nonumber\]Thus, \(h^{\prime}(a)=\cos(a^3) \cdot 3a^2\).
In other words, if \(h(x)=\sin(x^3)\), then \(h^{\prime}(x)=\cos(x^3) \cdot 3x^2\). Thus, if we think of \(h(x)=\sin(x^3)\) as the composition \((f \circ g)(x)=f \left(g(x)\right)\) where \(f(x)= \sin x\) and \(g(x)=x^3\), then the derivative of \(h(x)=\sin(x^3)\) is the product of the derivative of \(g(x)=x^3\) and the derivative of the function \(f(x)=\sin x\) evaluated at the function \(g(x)=x^3\). At this point, we anticipate that for \(h(x)=\sin\left(g(x)\right)\), it is quite likely that \(h^{\prime}(x)=\cos\left(g(x)\right)g^{\prime}(x)\). As we determined above, this is the case for \(h(x)=\sin(x^3)\).
Now that we have derived a special case of the Chain Rule, we state the general case and then apply it in a general form to other composite functions.
Let \(f\) and \(g\) be functions. For all \(x\) in the domain of \(g\) for which \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g(x)\), the derivative of the composite function\[h(x)=(f \circ g)(x)=f\left(g(x)\right)\nonumber\]is given by\[h^{\prime}(x)=f^{\prime}\left(g(x)\right)\cdot g^{\prime}(x).\nonumber\]Alternatively, if \(y\) is a function of \(u\), and \(u\) is a function of \(x\), then\[\dfrac{dy}{dx}=\dfrac{dy}{du} \cdot \dfrac{du}{dx}.\nonumber\]
- Proof (informal)
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At this point, we present a very informal proof of the Chain Rule. For simplicity's sake, we ignore certain issues: For example, we assume that \(g(x) \neq g(a)\) for \(x \neq a\) in some open interval containing \(a\). We begin by applying the limit definition of the derivative to the function \(h(x)\) to obtain \(h^{\prime}(a)\):\[h^{\prime}(a)=\lim_{x \to a}\dfrac{f\left(g(x)\right)-f\left(g(a)\right)}{x-a}.\nonumber\]Rewriting, we obtain\[h^{\prime}(a)=\lim_{x \to a}\dfrac{f\left(g(x)\right)-f\left(g(a)\right)}{g(x)-g(a)} \cdot \dfrac{g(x)-g(a)}{x-a}.\nonumber\]Although it is clear that\[\lim_{x \to a}\dfrac{g(x)-g(a)}{x-a}=g^{\prime}(a),\nonumber\]it is not obvious that\[\lim_{x \to a}\dfrac{f\left(g(x)\right)-f\left(g(a)\right)}{g(x)-g(a)}=f^{\prime}\left(g(a)\right).\nonumber\]To see that this is true, first recall that since \(g\) is differentiable at \(a\), \(g\) is also continuous at \(a\). Thus,\[\lim_{x \to a}g(x)=g(a).\nonumber\]Next, make the substitution \(y=g(x)\)and \(b=g(a)\) and use change of variables in the limit to obtain\[\lim_{x \to a}\dfrac{f\left(g(x)\right)-f \left(g(a) \right)}{g(x)-g(a)}=\lim_{y \to b}\dfrac{f(y)-f(b)}{y-b}=f^{\prime}(b)=f^{\prime}\left(g(a)\right).\nonumber\]Finally,\[h^{\prime}(a)=\lim_{x \to a}\dfrac{f\left(g(x)\right)-f\left(g(a)\right)}{g(x)-g(a)} \cdot \dfrac{g(x)-g(a)}{x-a}=f^{\prime}\left(g(a)\right)\cdot g^{\prime}(a).\nonumber\]
Q.E.D.
The Chain and Power Rules Combined
We can now apply the Chain Rule to composite functions, but note that we often need to use it with other rules. For example, to find derivatives of functions of the form \(h(x)=\left(g(x)\right)^n\), we need to use the Chain Rule combined with the Power Rule. To do so, we can think of \(h(x)=\left(g(x)\right)^n\) as \(f\left(g(x)\right)\) where \(f(x)=x^n\). Then \(f^{\prime}(x)=nx^{n-1}\). Thus, \(f^{\prime}\left(g(x)\right)=n\left(g(x)\right)^{n-1}\). This leads us to the derivative of a power function using the Chain Rule,\[h^{\prime}(x)=n\left(g(x)\right)^{n-1}\cdot g^{\prime}(x)\nonumber\]
For all values of \(x\) for which the derivative is defined, if\[h(x)=\left(g(x)\right)^n,\nonumber\]Then\[h^{\prime}(x)=n\left(g(x)\right)^{n-1}\cdot g^{\prime}(x) \label{genpow}.\]
Find the derivative of \(h(x)=\frac{1}{(3x^2+1)^2}\).
- Solution
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First, rewrite \(h(x)=\frac{1}{(3x^2+1)^2}=(3x^2+1)^{-2}\). Applying the Power Rule with \(g(x)=3x^2+1\), we have\[h^{\prime}(x)=-2(3x^2+1)^{-3}\cdot 6x.\nonumber\]Rewriting back to the original form gives us\[h^{\prime}(x)=\dfrac{-12x}{(3x^2+1)^3}.\nonumber\]
Find the derivative of \(h(x)=(2x^3+2x-1)^4\).
- Answer
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\(h^{\prime}(x)=4(2x^3+2x-1)^3(6x^2+2)=8(3x^2+1)(2x^3+2x-1)^3\)
Find the derivative of \(h(x)=\sin^3x\).
- Solution
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First recall that \(\sin^3x=(\sin x)^3\), so we can rewrite \(h(x)=\sin^3x\) as \(h(x)=(\sin x)^3\).
Applying the Power Rule with \(g(x)=\sin x\), we obtain\[h^{\prime}(x)=3(\sin x)^2\cos x=3\sin^2x\cos x.\nonumber\]
Find the equation of a line tangent to the graph of \(h(x)=\frac{1}{(3x-5)^2}\) at \(x=2\).
- Solution
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Because we are finding an equation of a line, we need a point. The \(x\)-coordinate of the point is \(2\). To find the \(y\)-coordinate, substitute \(2\) into \(h(x)\). Since \(h(2)=\frac{1}{(3(2)-5)^2}=1\), the point is \((2,1)\).
For the slope, we need \(h^{\prime}(2)\). To find \(h^{\prime}(x)\), first we rewrite \(h(x)=(3x-5)^{-2}\) and apply the Power Rule to obtain\[h^{\prime}(x)=-2(3x-5)^{-3}(3)=-6(3x-5)^{-3}.\nonumber\]By substituting, we have \(h^{\prime}(2)=-6(3(2)-5)^{-3}=-6\).
Therefore, the line has equation \(y-1=-6(x-2)\). Rewriting, the equation of the line is \(y=-6x+13\).
Find the equation of the line tangent to the graph of \(f(x)=(x^2-2)^3\) at \(x=-2\).
- Answer
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\(y=-48x-88\)
Combining the Chain Rule with Other Rules
Now that we can combine the Chain Rule and the Power Rule, we examine how to combine the Chain Rule with the other rules we have learned. In particular, we can use it with the formulas for the derivatives of trigonometric functions or with the Product Rule.
Find the derivative of \(h(x)=\cos\left(g(x)\right)\).
- Solution
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Think of \(h(x)=\cos\left(g(x)\right)\) as \(f\left(g(x)\right)\) where \(f(x)=\cos x\). Since \(f^{\prime}(x)=-\sin x\), we have \(f^{\prime}\left(g(x)\right)=-\sin\left(g(x)\right)\). Then we do the following calculation.\[\begin{array}{rclcl} h^{\prime}(x) & = & f^{\prime}\left(g(x)\right)\cdot g^{\prime}(x) & \quad & \left( \text{Chain Rule} \right) \\ & = & -\sin\left(g(x)\right)\cdot g^{\prime}(x) & \quad & \left( \text{substituting }f^{\prime}\left(g(x)\right) =-\sin\left(g(x)\right) \right) \\ \end{array}\nonumber\]Thus, the derivative of \(h(x)=\cos\left(g(x)\right)\) is given by \(h^{\prime}(x)=-\sin\left(g(x)\right)\cdot g^{\prime}(x)\).
In the following example, we apply the rule that we have just derived.
Find the derivative of \(h(x)=\cos(5x^2)\).
- Solution
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Let \(g(x)=5x^2\). Then \(g^{\prime}(x)=10x\). Using the result from the previous example,\[h^{\prime}(x)=-\sin(5x^2) \cdot 10x=-10x\sin(5x^2).\nonumber\]
Find the derivative of \(h(x)=\sec{(4x^5+2x)}\).
- Solution
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Apply the Chain Rule to \(h(x)=\sec{\left(g(x)\right)}\) to obtain\[h^{\prime}(x)=\sec{(g(x))}\tan\left(g(x)\right)\cdot g^{\prime}(x).\nonumber\]In this problem, \(g(x)=4x^5+2x\), so we have \(g^{\prime}(x)=20x^4+2\). Therefore, we obtain\[h^{\prime}(x)=\sec{(4x^5+2x)}\tan(4x^5+2x)(20x^4+2)=(20x^4+2)\sec{(4x^5+2x)}\tan(4x^5+2x).\nonumber\]
Find the derivative of \(h(x)=\sin(7x+2)\).
- Answer
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\(h^{\prime}(x)=7\cos(7x+2)\)
Find the derivative of \(h(x)=(2x+1)^5(3x-2)^7\).
- Solution
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First apply the Product Rule, then apply the Chain Rule to each term of the product.\[\begin{array}{rclcl} h^{\prime}(x) & = & \dfrac{d}{dx}\left((2x+1)^5\right) \cdot (3x-2)^7+\dfrac{d}{dx}\left((3x-2)^7\right) \cdot (2x+1)^5 & \quad & \left( \text{Product Rule} \right) \\[12pt] & = & 5(2x+1)^4 \cdot 2 \cdot (3x-2)^7+7(3x-2)^6 \cdot 3 \cdot (2x+1)^5 & \quad & \left( \text{Chain Rule} \right) \\[12pt] & = & 10(2x+1)^4(3x-2)^7+21(3x-2)^6(2x+1)^5 & \quad & \left( \text{simplifying} \right) \\[12pt] & = & (2x+1)^4(3x-2)^6(10(3x-2)+21(2x+1)) & \quad & \left( \text{factoring} \right) \\[12pt] & = & (2x+1)^4(3x-2)^6(72x+1) & \quad & \left( \text{simplifying} \right) \\[12pt] \end{array}\nonumber\]
Find the derivative of \(h(x)=\frac{x}{(2x+3)^3}\).
- Answer
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\(h^{\prime}(x)=\frac{3-4x}{(2x+3)^4}\)
Composites of Three or More Functions
We can now combine the Chain Rule with other rules for differentiating functions, but when differentiating the composition of three or more functions, we need to apply the Chain Rule more than once. Let's look at this situation in general terms. We can generate a formula without remembering it, as we can apply the Chain Rule multiple times.
In general terms, first we let\[k(x)=h\left(f\left(g(x)\right)\right).\nonumber\]Then, applying the Chain Rule once we obtain\[k^{\prime}(x)=\dfrac{d}{dx}\left(h\left(f\left(g(x)\right)\right)\right)=h^{\prime}\left(f\left(g(x)\right)\right) \cdot \dfrac{d}{dx}\left(f\left(g(x)\right)\right).\nonumber\]Applying the Chain Rule again, we obtain\[k^{\prime}(x)=h^{\prime}\left(f\left(g(x)\right)\right)\cdot f^{\prime}\left(g(x)\right)\cdot g^{\prime}(x).\nonumber\]
For all values of \(x\) for which the function is differentiable, if\[k(x)=h\left(f\left(g(x)\right)\right),\nonumber\]then\[k^{\prime}(x)=h^{\prime}\left(f\left(g(x)\right)\right)\cdot f^{\prime}\left(g(x)\right)\cdot g^{\prime}(x).\nonumber\]In other words, we are applying the Chain Rule twice.
Notice that the derivative of the composition of three functions has three parts. (Similarly, the derivative of the composition of four functions has four parts, and so on.) Also, remember, we can always work from the outside in, taking one derivative at a time.
Find the derivative of \(k(x)=\cos^4(7x^2+1)\).
- Solution
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First, rewrite \(k(x)\) as\[k(x)=\left(\cos(7x^2+1)\right)^4.\nonumber\]Then apply the Chain Rule several times.\[\begin{array}{rclcl} k^{\prime}(x) & = & 4(\cos(7x^2+1))^3\cdot\dfrac{d}{dx}\left(\cos(7x^2+1)\right) & \quad & \left( \text{Chain Rule} \right) \\[16pt] & = & 4(\cos(7x^2+1))^3(-\sin(7x^2+1))\cdot\dfrac{d}{dx}\left(7x^2+1\right) & \quad & \left( \text{Chain Rule} \right) \\[16pt] & = & 4(\cos(7x^2+1))^3(-\sin(7x^2+1))(14x) & \quad & \left( \text{Chain Rule} \right) \\[16pt] & = & -56x\sin(7x^2+1)\cos^3(7x^2+1) & \quad & \left( \text{simplifying} \right) \\[16pt] \end{array}\nonumber\]
Find the derivative of \(h(x)=\sin^6(x^3)\).
- Answer
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\(h^{\prime}(x)=18x^2\sin^5(x^3)\cos(x^3)\)
A particle moves along a coordinate axis. Its position at time t is given by \(s(t)=\sin(2t)+\cos(3t)\). What is the velocity of the particle at time \(t=\frac{\pi}{6}\)?
- Solution
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To find \(v(t)\), the velocity of the particle at time \(t\), we must differentiate \(s(t)\). Thus,\[v(t)=s^{\prime}(t)=2\cos(2t)-3\sin(3t).\nonumber\]
Let \(h(x)=f\left(g(x)\right)\). If \(g(1)=4,g^{\prime}(1)=3\), and \(f^{\prime}(4)=7\), find \(h^{\prime}(1)\).
- Solution
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Use the Chain Rule, then substitute.\[\begin{array}{rclcl} h^{\prime}(1) & = & f^{\prime}\left(g(1)\right)\cdot g^{\prime}(1) & \quad & \left( \text{Chain Rule} \right) \\ & = & f^{\prime}(4) \cdot 3 & \quad & \left( \text{substituting } g(1)=4 \text{ and } g^{\prime}(1)=3 \right) \\ & = & 7 \cdot 3 & \quad & \left( \text{substituting }f^{\prime}(4)=7 \right) \\ & = & 21 & \quad & \left( \text{simplifying} \right) \\ \end{array}\nonumber\]
Given \(h(x)=f(g(x))\). If \(g(2)=-3,g^{\prime}(2)=4\), and \(f^{\prime}(-3)=7\), find \(h^{\prime}(2)\).
- Answer
-
28
The Chain Rule Using Leibniz's Notation
As with other derivatives that we have seen, we can express the Chain Rule using Leibniz's notation. This notation for the Chain Rule is used heavily in physics applications.
For \(h(x)=f(g(x))\), let \(u=g(x)\) and \(y=h(x)=f(u)\). Thus,\[h^{\prime}(x)=\dfrac{dy}{dx}\nonumber\]\[f^{\prime}(g(x))=f^{\prime}(u)=\dfrac{dy}{du}\nonumber\]and\[g^{\prime}(x)=\dfrac{du}{dx}.\nonumber\]Consequently,\[\dfrac{dy}{dx}=h^{\prime}(x)=f^{\prime}\left(g(x)\right)\cdot g^{\prime}(x)=\dfrac{dy}{du} \cdot \dfrac{du}{dx}.\nonumber\]
If \(y\) is a function of \(u\), and \(u\) is a function of \(x\), then\[\dfrac{dy}{dx}=\dfrac{dy}{du} \cdot \dfrac{du}{dx}.\nonumber\]
Find the derivative of \(y=\left(\frac{x}{3x+2}\right)^5\).
- Solution
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First, let \(u=\frac{x}{3x+2}\). Thus, \(y=u^5\). Next, find \(\frac{du}{dx}\) and \(\frac{dy}{du}\). Using the Quotient Rule,\[\dfrac{du}{dx}=\dfrac{2}{(3x+2)^2}\nonumber\]and\[\dfrac{dy}{du}=5u^4,\nonumber\]Finally, we put it all together.\[\begin{array}{rclcl} \dfrac{dy}{dx} & = & \dfrac{dy}{du} \cdot \dfrac{du}{dx} & \quad & \left( \text{Chain Rule} \right) \\[16pt] & = & 5u^4 \cdot \dfrac{2}{(3x+2)^2} & \quad & \left( \text{substituting }\dfrac{dy}{du}=5u^4\text{ and }\dfrac{du}{dx}=\dfrac{2}{(3x+2)^2}\right) \\[16pt] & = & 5\left(\dfrac{x}{3x+2}\right)^4 \cdot \dfrac{2}{(3x+2)^2} & \quad & \left( \text{substituting }u=\dfrac{x}{3x+2} \right) \\[16pt] & = & \dfrac{10x^4}{(3x+2)^6} & \quad & \left( \text{simplifying} \right) \\[16pt] \end{array}\nonumber\]It is important to remember that when using the Leibniz form of the Chain Rule, the final answer must be expressed entirely in terms of the original variable given in the problem.
Find the derivative of \(y=\tan(4x^2-3x+1)\).
- Solution
-
First, let \(u=4x^2-3x+1\). Then \(y=\tan u\). Next, find \(\frac{du}{dx}\) and \(\frac{dy}{du}\):\[\dfrac{du}{dx}=8x-3 \quad \text{and} \quad \dfrac{dy}{du}=\sec^2{u}.\nonumber\]Finally, we put it all together.\[\begin{array}{rclcl} \dfrac{dy}{dx} & = & \dfrac{dy}{du} \cdot \dfrac{du}{dx} & \quad & \left( \text{Chain Rule} \right) \\[16pt] & = & \sec^2{u} \cdot (8x-3) & \quad & \left( \dfrac{du}{dx}=8x-3\text{ and }\dfrac{dy}{du}=\sec^2{u} \right) \\[16pt] & = & \sec^2{(4x^2-3x+1)} \cdot (8x-3) & \quad & \left( \text{substituting }u=4x^2-3x+1. \right) \\[16pt] \end{array}\nonumber\]
Use Leibniz's notation to find the derivative of \(y=\cos(x^3)\). Ensure that the final answer is expressed entirely in terms of the variable \(x\).
- Answer
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\(\frac{dy}{dx}=-3x^2\sin(x^3)\).