Objectives
- State and use the chain rule using different notations
- Use the chain rule to compute derivatives in conjunction with the derivative rules of Section 2.3 using algebraic, graphical, or numerical representations
Section 2.5 The Chain Rule
Subsection 2.5.1 Before Class
https://mymedia.ou.edu/media/2.5-1/1_wrjdpiy3Subsubsection 2.5.1.1 Review: Composition of Functions
Example 2.5.1.
If \(f(u)= \sqrt{u}\) and \(u(x) = x^2 + 1\text{,}\) find the composition \((f\circ u)(x)\text{.}\)
Solution.
\((f\circ u)(x) = \sqrt{x^2+1}\)
Example 2.5.2.
If \(f(x) = \sqrt{x}\) and \(g(x) = \sqrt{2-x}\text{,}\) find- \(\displaystyle f\circ g\)
- \(\displaystyle g\circ f\)
- \(\displaystyle f\circ f\)
- \(\displaystyle g\circ g\)
Solution.
- \(\displaystyle f\circ g = \sqrt{\sqrt{2-x}}\)
- \(\displaystyle g\circ f = \sqrt{2-\sqrt{x}}\)
- \(\displaystyle f\circ f = \sqrt{\sqrt{x}}\)
- \(\displaystyle g\circ g = \sqrt{2-\sqrt{2-x}}\)
Example 2.5.3.
Let \(k(x) = \cos 2x\text{.}\) Find \(f,g\) such that \(k(x) = f(g(x))\text{.}\)
Solution.
\(f(x) = \cos x\) and \(g(x) = 2x\)
Example 2.5.4.
Let \(f(x) = \sec^2 (x^2 + 9)\text{.}\) Find functions \(a,b,c\) such that \(f(x) = (a\circ b\circ c)(x)\)
Solution.
\(a(x) = x^2\text{,}\) \(b(x) = \sec x\text{,}\) and \(c(x) = x^2+9\)
Subsubsection 2.5.1.2 The Chain Rule
The chain rule relies on being able to decompose a function into smaller pieces, and doing things in the right order.
Theorem 2.5.5. The Chain Rule.
If \(g\) is differentiable at \(x\text{,}\) and \(f\) is differentiable at \(g(x)\text{,}\) then the composite function \(F = f\circ g\) defined by \(F(x) = f(g(x))\) is differentiable at \(x\text{,}\) and \(F'(x) = f'(g(x))\cdot g'(x)\)
In Leibniz notatation, if \(y = f(u)\) and \(u = g(x)\) are both differentiable functions, then \(\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot \dfrac{du}{dx}\)
Example 2.5.6.
Let \(k(x) = \cos 2x\text{.}\) Find \(k'(x)\) using Example 2.5.3.
Solution.
\(k'(x) = -2\sin 2x\)
Example 2.5.7.
Let \(g(x) = \sin 4x\text{.}\) Find \(g'(x)\text{.}\)
Solution.
\(g'(x) = 4\cos 4x\)
Example 2.5.8.
Let \(f(x) = \sec^2(x^2+9)\text{.}\) Find \(f'(x)\) using Example 2.5.4.
Solution.
\(f'(x) = 2\sec(x^2+9)\cdot \sec(x^2+9)\tan(x^2+9)\cdot 2x\)
Subsection 2.5.2 Pre-Class Activities
Example 2.5.9.
You are given a composite function. Identify the inner function \(u = g(x)\) and the outer function \(y = f(u)\text{.}\)- \(\displaystyle \sqrt[3]{1+4x}\)
- \(\displaystyle \sin (\cot x)\)
- \(\displaystyle (5x^6 + 2x^3)^4\)
Solution.
- \(y = f(u) = \sqrt[3]{u}\) and \(u = g(x) = 1+4x\)
- \(y = f(u) = \sin u\) and \(u = g(x) = \cot x\)
- \(y = f(u) = u^4\) and \(u = g(x) = 5x^6 + 2x^3\)
Example 2.5.10.
Set \(h(x) = \sin (x^2)\text{.}\) Identify the inner function \(u = g(x)\) and the outer function \(y = f(u)\text{.}\) Then, find the derivative \(h'(x)\text{.}\)
Solution.
\(u = g(x) = x^2\) and \(y = f(u) = \sin u\text{;}\) \(f'(x) = 2x\cos(x^2)\)
Example 2.5.11.
Set \(k(t) = \sin^2(t)\text{.}\) Identify the inner function \(u = g(t)\) and the outer function \(y = f(u)\text{.}\) Then, find the derivative \(k'(t)\text{.}\)
Solution.
\(u = g(t) = \sin t\) and \(y = f(u) = u^2\text{;}\) \(k'(t) = 2\sin t\cos t\)
Subsection 2.5.3 In Class
Question 2.5.12.
If \(h(x) = f(g(x))\text{,}\) use the chain rule to write \(h'(x)\) in prime notation and Leibniz notation.
Solution.
In prime notation, \(h'(x) = f'(g(x))\cdot g'(x)\text{.}\) In Leibniz notation, \(\dfrac{dh}{dx} = \dfrac{df}{dg}\cdot \dfrac{dg}{dx}\)
Example 2.5.13.
Let \(f(x) = \sqrt{x^2 + 1}\text{.}\) Find \(f'(x)\text{.}\)
Solution.
\(f'(x) = \dfrac{1}{2}(x^2+1)^{-1/2}\cdot 2x\)
Example 2.5.14.
Let \(g(x) = \cos (x^2)\text{.}\) Find \(g'(x)\text{.}\)
Solution.
\(g'(x) = -2x\sin(2x)\)
Example 2.5.15.
Find the derivative of \(k(t) = (2t+1)^5(t^3-t+1)^4\)
Solution.
\(k'(t) = 5(2t+1)^4(2)(t^3-t+1)^4 + 4(t^3-t+1)^3(3t^2-1)(2t+1)^5\)
Example 2.5.16.
Let \(h(x) = \sin^2(\sqrt{x^2-1})\text{.}\) Find \(h'(x)\text{.}\)
Solution.
\(h'(x) = 2\sin\sqrt{x^2-1}\cdot \cos\sqrt{x^2-1}\cdot \dfrac{1}{2}(x^2-1)^{-1/2}\cdot 2x\)
Example 2.5.17.
Find the first derivative of \(F(x) = (5x^5+2x^3)^4\)
Solution.
\(F'(x) = 4(5x^5+2x^3)^3(25x^4+6x^2)\)
Example 2.5.18.
Find the first derivative of \(h(t) = (2-\sin t)^{3/2}\)
Solution.
\(h'(t) = \dfrac{3}{2}(2-\sin t)^{1/2}\cdot (-\cos t)\)
Example 2.5.19.
Find the first derivative of \(y=\dfrac{1}{(\cos t + \tan t)^2}\)
Solution.
\(\dfrac{dy}{dx} = -2(\cos t + \tan t)^{-3}(-\sin t + \sec^2t)\)
Example 2.5.20.
Find the first derivative of \(h(\theta) = \tan (\theta^2\sin\theta)\)
Solution.
\(h'(\theta) = \sec^2(\theta^2\sin\theta)\cdot (2\theta\sin\theta + \theta^2\cos\theta)\)
Example 2.5.21.
Find the first derivative of \(y=\lrpar{\dfrac{1-\cos 2x}{1+\sin 2x}}^3\)
Solution.
\(\dfrac{dy}{dx} = 3\lrpar{\dfrac{1-\cos 2x}{1+\sin 2x}}^2\lrpar{\dfrac{(1+\sin2x)(2\sin2x)-(1-\cos2x)(2\cos 2x)}{(1+\sin 2x)^2}}\)
Example 2.5.22.
Find the first derivative of \(f(t) = \sqrt{t+\sqrt{t}}\)
Solution.
\(f'(t) = \dfrac{1}{2}(t+t^{1/2})^{-1/2}\lrpar{1+\dfrac{1}{2}t^{-1/2}}\)
Example 2.5.23.
Find the first derivative of \(r(x) = (x^2+1)^3(x^2+2)^6\)
Solution.
\(r'(x) = 3(x^2+1)^2(2x)(x^2+2)^6 + 6(x^2+2)(2x)(x^2+1)^3\)
Example 2.5.24.
Find the first derivative of \(y = \sqrt[5]{\dfrac{x}{x-1}}\)
Solution.
\(\dfrac{dy}{dx} = \dfrac{1}{5}\lrpar{\dfrac{x}{x-1}}^{-4/5}\lrpar{\dfrac{-1}{(x-1)^2}}\)
Example 2.5.25.
Find the first derivative of \(z = \sqrt{\sin(1+x^2)}\)
Solution.
\(\dfrac{dz}{dx} = \dfrac{1}{2}(\sin(1+x^2))^{-1/2}\cdot \cos(1+x^2)(2x)\)
Example 2.5.26.
Find the first derivative of \(A(t) = \dfrac{t^2}{\sqrt{t^3+1}}\)
Solution.
\(A'(t) = \dfrac{(2t)(\sqrt{t^3+1})-(t^2)\lrpar{\dfrac{1}{2}(t^3+1)^{-1/2}(3t^2)}}{t^3+1}\)
Example 2.5.27.
Find the first derivative of \(f(x) = \cos^4(\tan^3(x))\)
Solution.
\(f'(x)=4\cos^3(\tan^3x)\cdot -\sin(\tan^3x)\cdot 3\tan^2x\cdot\sec^2x\)
Example 2.5.28.
Find the first derivative of \(y = \cos\sqrt{\sin(\tan\pi x)}\)
Solution.
\(\dfrac{dy}{dx} = -\sin\sqrt{\sin(\tan\pi x)}\cdot \dfrac{1}{2}(\sin(\tan\pi x))^{-1/2}(\cos(\tan\pi x))\cdot \sec^2(\pi x)\cdot \pi\)
Example 2.5.29.
Find an equation of the tangent line to the curve \(y = \sqrt{1+x^3}\) at the point \((2,3)\)
Solution.
\(y = 2x-1\)
Example 2.5.30.
Let \(f(x) = [g(\cos x)]^2\text{.}\) Write an expression for \(f'(x)\text{.}\)
Solution.
\(f'(x) = 2g(\cos x)\cdot g'(\cos x)\cdot -\sin x\)
Example 2.5.31.
If \(h(x) = \sqrt{4+3f(x)}\text{,}\) \(f(1) = 7\text{,}\) and \(f'(1) = 4\text{,}\) find \(h'(1)\text{.}\)
Solution.
\(h'(1) = \dfrac{6}{5}\)
Example 2.5.32.
If \(g(x) = \sqrt{f(x)}\text{,}\) where \(f\) is the function shown, evaluate \(g'(3)\text{.}\)
Solution.
\(g'(3) = -\dfrac{1}{2\sqrt{3}}\)
Subsection 2.5.4 After Class Activities
Example 2.5.33.
Find the first and second derivatives of \(y = \sin (\cos 4\theta)\)
Solution.
\(\dfrac{dy}{dx} = -4\sin(4\theta)\cdot cos(\cos 4\theta)\) and \(\dfrac{d^2y}{dx^2} = -16\cos(4\theta)\cdot \cos(\cos 4\theta) -16\sin^2(4\theta)\cdot \sin (\cos 4\theta)\)
Example 2.5.34.
Find the first and second derivatives of \(y = \dfrac{4x}{\sqrt{x+1}}\)
Solution.
\(\dfrac{dy}{dx} = \dfrac{2x+4}{(x+1)^{3/2}}\) and \(\dfrac{d^2y}{dx^2} = \dfrac{2(x+1)^{3/2}-(2x+4)\lrpar{\dfrac{3}{2}(x+1)^{1/2}}}{(x+1)^3}\)
Example 2.5.35.
Find \(D^{35} \sin \pi x\)
Solution.
\(D^{35}[\sin \pi x] = -\pi^{35}\cos\pi x\)
Example 2.5.36.
A table of values for \(f,g,g'\) and \(g'\) is given:| \(x\) | \(f(x)\) | \(g(x)\) | \(f'(x)\) | \(g'(x)\) |
| \(1\) | \(3\) | \(2\) | \(4\) | \(6\) |
| \(2\) | \(1\) | \(8\) | \(5\) | \(7\) |
| \(3\) | \(7\) | \(2\) | \(7\) | \(9\) |
- Find \(h'(1)\text{,}\) if \(h(x) = f(g(x))\)
- Find \(H'(1)\text{,}\) if \(H(x) = g(f(x))\)
Solution.
- Find \(h'(1) = 30\)
- Find \(H'(1) = 36\)
Example 2.5.37.
Suppose \(f\) is differentiable on \(\R\text{,}\) and \(A\) is a real number. Let \(F(x) = f(x^A)\) and \(G(x) = [f(x)]^A\text{.}\) Find expressions for \(F'(x)\) and \(G'(x)\text{.}\)
Solution.
\(F'(x) = f'(x^A)\cdot Ax^{A-1}\) and \(G'(x) = A[f(x)]^{A-1}\cdot f'(x)\)
Example 2.5.38.
The chain rule is often a source of confusion and frustration for students in (and beyond) calculus. What do you think will help the chain rule stick out in your mind?
Solution.
Answers vary
