When working with composite functions where function values are given in tables rather than formulas, we can still apply the chain rule using the tabulated data.
Graph Description: A piecewise linear graph of function \(g(x)\) showing connected line segments. The graph starts at the origin, rises to a peak at point \((1, 4)\) with slope 2, then descends to point \((2, 2)\), and continues descending with slope -4 to reach \((3, 0)\). Key values include \(g(1) = 2\), \(g(2) = 4\), and \(g(3) = 0\). The slopes at various points represent \(g'(x)\) values, with \(g'(3) = -4\) indicated.
Given the following function values and derivatives:
| \(x\) | 1 | 2 | 3 |
|---|---|---|---|
| \(f(x)\) | 3 | 2 | 1 |
| \(f'(x)\) | 4 | 5 | 6 |
From the graph: \(g(1) = 2\), \(g(2) = 4\), \(g(3) = 0\), \(g'(1) = 2\), \(g'(2)\) = DNE, \(g'(3) = -4\)
Define \(h(x) = f(g(x))\) and \(k(x) = g(f(x))\). Find \(h'(1)\) and \(k'(3)\).
Solution for \(h'(1)\):
Using the chain rule: \(h'(x) = f'(g(x)) \cdot g'(x)\)
\[h'(1) = f'(g(1)) \cdot g'(1)\]From the table and graph: \(g(1) = 2\), so \(f'(g(1)) = f'(2) = 5\)
Also, \(g'(1) = 2\)
\[h'(1) = f'(2) \cdot g'(1) = 5 \cdot 2 = 10\]Solution for \(k'(3)\):
Using the chain rule: \(k'(x) = g'(f(x)) \cdot f'(x)\)
\[k'(3) = g'(f(3)) \cdot f'(3)\]From the table: \(f(3) = 1\), so \(g'(f(3)) = g'(1) = 2\)
Also, \(f'(3) = 6\)
\[k'(3) = g'(1) \cdot f'(3) = 2 \cdot 6 = 12\]Find \(\frac{dy}{dx}\) for \(y = e^{5x^2 - 8}\)
Solution:
Let \(u = 5x^2 - 8\) (inside function)
\(\frac{du}{dx} = 10x\)
Applying the chain rule:
\[\frac{dy}{dx} = \frac{d}{dx}[e^u] = \frac{de^u}{du} \cdot \frac{du}{dx}\] \[= e^u \cdot \frac{du}{dx}\] \[= e^{5x^2 - 8} \cdot 10x\]Find \(\frac{dy}{dx}\) for \(y = \sin(e^{3x})\)
Solution:
This requires applying the chain rule twice.
Let \(u = e^{3x}\) (inside function)
\[\frac{dy}{dx} = \frac{d}{dx}[\sin u] = \frac{d\sin u}{du} \cdot \frac{du}{dx}\] \[= \cos u \cdot \frac{du}{dx}\] \[= \cos(e^{3x}) \cdot \frac{d}{dx}[e^{3x}]\]Now apply the chain rule again to \(\frac{d}{dx}[e^{3x}]\):
Let \(v = 3x\), so \(\frac{dv}{dx} = 3\)
\[\frac{d}{dx}[e^{3x}] = \frac{de^v}{dv} \cdot \frac{dv}{dx} = e^v \cdot 3 = e^{3x} \cdot 3\]Therefore:
\[\frac{dy}{dx} = \cos(e^{3x}) \cdot e^{3x} \cdot 3\]Find \(\frac{dy}{dx}\) for \(y = \csc(\cos(x^3 + 5))\)
Solution:
This requires multiple applications of the chain rule.
Let \(u = \cos(x^3 + 5)\)
\[\frac{dy}{dx} = \frac{d}{dx}[\csc u] = \frac{d\csc u}{du} \cdot \frac{du}{dx}\] \[= -\csc u \cot u \cdot \frac{du}{dx}\] \[= -\csc(\cos(x^3 + 5)) \cot(\cos(x^3 + 5)) \cdot \frac{d}{dx}[\cos(x^3 + 5)]\]Now find \(\frac{d}{dx}[\cos(x^3 + 5)]\):
Let \(v = x^3 + 5\), so \(\frac{dv}{dx} = 3x^2\)
\[\frac{d}{dx}[\cos v] = \frac{d\cos v}{dv} \cdot \frac{dv}{dx} = -\sin v \cdot 3x^2 = -\sin(x^3 + 5) \cdot 3x^2\]Putting it all together:
\[\frac{dy}{dx} = -\csc(\cos(x^3 + 5)) \cot(\cos(x^3 + 5)) \cdot (-\sin(x^3 + 5)) \cdot 3x^2\] \[= \csc(\cos(x^3 + 5)) \cot(\cos(x^3 + 5)) \cdot \sin(x^3 + 5) \cdot 3x^2\]Extended Chain Rule for Nested Compositions:
For a composition of three functions \(f(g(h(x)))\):
\[\frac{d}{dx}[f(g(h(x)))] = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)\]In Leibniz notation, with \(u = g(h(x))\):
\[\frac{d}{dx}[f(u)] = \frac{df}{du} \cdot \frac{du}{dx}\]Find \(\frac{dy}{dx}\) for \(y = [\sec(x^2) \cdot (x^2 + 9)]^{3/4}\)
Solution:
Let \(u = \sec(x^2) \cdot (x^2 + 9)\)
Applying the chain rule:
\[\frac{dy}{dx} = \frac{d}{dx}[u^{3/4}] = \frac{3}{4}u^{-1/4} \cdot \frac{du}{dx}\] \[= \frac{3}{4}[\sec(x^2)(x^2 + 9)]^{-1/4} \cdot \frac{d}{dx}[\sec(x^2)(x^2 + 9)]\]Now apply the product rule to find \(\frac{du}{dx}\):
\[\frac{d}{dx}[\sec(x^2)(x^2 + 9)] = \frac{d}{dx}[\sec(x^2)] \cdot (x^2 + 9) + \sec(x^2) \cdot \frac{d}{dx}[x^2 + 9]\]For \(\frac{d}{dx}[\sec(x^2)]\), use the chain rule with \(v = x^2\):
\[\frac{d}{dx}[\sec v] = \sec v \tan v \cdot \frac{dv}{dx} = \sec(x^2)\tan(x^2) \cdot 2x\]Therefore:
\[\frac{du}{dx} = \sec(x^2)\tan(x^2) \cdot 2x \cdot (x^2 + 9) + \sec(x^2) \cdot 3x^2\]Final answer:
\[\frac{dy}{dx} = \frac{3}{4}[\sec(x^2)(x^2 + 9)]^{-1/4} \left[\sec(x^2)\tan(x^2) \cdot 2x \cdot (x^2 + 9) + \sec(x^2) \cdot 3x^2\right]\]Find \(\frac{dy}{dx}\) for \(y = \frac{\tan(x^3 + 5)}{e^{\sin x}}\)
Solution:
Let \(N = \tan(x^3 + 5)\) and \(D = e^{\sin x}\)
Using the quotient rule:
\[y' = \frac{N'D - ND'}{D^2}\]Find \(N'\) using the chain rule:
\[N' = \frac{d}{dx}[\tan(x^3 + 5)] = \sec^2(x^3 + 5) \cdot 3x^2\]Find \(D'\) using the chain rule:
\[D' = \frac{d}{dx}[e^{\sin x}] = e^{\sin x} \cdot \cos x\]Therefore:
\[y' = \frac{\sec^2(x^3 + 5) \cdot 3x^2 \cdot e^{\sin x} - \tan(x^3 + 5) \cdot e^{\sin x} \cos x}{(e^{\sin x})^2}\] \[= \frac{e^{\sin x}[\sec^2(x^3 + 5) \cdot 3x^2 - \tan(x^3 + 5) \cos x]}{e^{2\sin x}}\] \[= \frac{\sec^2(x^3 + 5) \cdot 3x^2 - \tan(x^3 + 5) \cos x}{e^{\sin x}}\]Given \(g(x) = \cos(\pi + f(x))\), where \(f(1) = \frac{1}{2}\) and \(f'(1) = 3\), find \(g'(1)\).
Solution:
Using the chain rule:
\[g'(x) = \frac{d}{dx}[\cos(\pi + f(x))]\]Let \(u = \pi + f(x)\), so \(\frac{du}{dx} = f'(x)\)
\[g'(x) = \frac{d\cos u}{du} \cdot \frac{du}{dx} = -\sin u \cdot f'(x)\] \[= -\sin(\pi + f(x)) \cdot f'(x)\]Evaluate at \(x = 1\):
\[g'(1) = -\sin(\pi + f(1)) \cdot f'(1)\] \[= -\sin\left(\pi + \frac{1}{2}\right) \cdot 3\] \[= -\sin\left(\frac{3\pi}{2}\right) \cdot 3\] \[= -(-1) \cdot 3 = 3\]Alternatively, using the product-to-sum identity \(\pi \frac{d}{dv}v^2 = \pi \frac{dv^2}{dv} \cdot \frac{dv}{dx}\):
\[= -\pi(2v) \cdot v'\] \[= -\pi \cdot 2f(x) \cdot f'(x)\] \[g'(1) = -\pi \cdot 2 \cdot \frac{1}{2} \cdot 3 = 2\pi f(x) \cdot f'(x)\]Key Takeaway: When applying the chain rule multiple times, work from the outside in. Identify the outermost function first, then systematically work through each nested function, multiplying the derivatives as you go.