Section 3.6 The Chain Rule
The chain rule vastly expands the complexity of functions that can be differentiated.
Fact 3.6.1. Chain Rule.
If \(f(x)\) is differentiable at the point \(u=g(x)\) and \(g(x)\) is differentiable at \(x\text{,}\) then the composite function \((f\circ g)(x)=f(g(x))\) is differentiable at \(x\) and
Alternately, if \(y=f(u)\) and \(u=g(x)\text{,}\) then
where \(\frac{dy}{du}\) is evaluated at \(u=f(x)\text{.}\)
Example 3.6.2.
Let \(y=\cos(x^2+3x-1)\text{.}\) Find \(y'\text{.}\)
\(y'= -(2x+3)\sin(x^2+3x-1)\)
Example 3.6.3.
\(\displaystyle f(x)=e^{\sqrt{x}}\text{;}\) find \(f'(x)\)
\(\displaystyle s = (5t^2 + \sin t)^3\text{;}\) find \(\frac{ds}{dt}\)
\(\displaystyle g(x) = \frac{2x-1}{x+\sqrt{x^2+1}}\text{;}\) find \(g'(x)\)
\(\displaystyle y = \sin^3 \left(\sqrt{1-x}\right)\text{;}\) find \(\frac{dy}{dx}\)
Find \(\displaystyle \frac{d}{dx} |x| = \frac{d}{dx} \sqrt{x^2}\)
\(r = \sin\left(\theta^2\right) \cos (2\theta)\text{;}\) find \(r'\)
\(\displaystyle \displaystyle f'(x) = \frac{e^{\sqrt{x}}}{2\sqrt{x}}\)
\(\displaystyle \displaystyle s' = 3(5t^2+\sin t)^2 (10t+\cos t)\)
\(\displaystyle \displaystyle g'(x) = \frac{2-x-\sqrt{x^2+1}}{\sqrt{x^2+1}\left(x+\sqrt{x^2+1}\right)^2}\)
\(\displaystyle \displaystyle y' = \frac{-3\sin^2 \left(\sqrt{1-x}\right) \cos \left(\sqrt{1-x}\right)}{2\sqrt{1-x}}\)
\(\displaystyle \displaystyle \frac{d}{dx} |x| = \frac{x}{|x|}\)
\(\displaystyle \displaystyle r' = 2\theta \cos(2\theta) \cos(\theta^2) - 2\sin(2\theta)\sin(\theta^2)\)