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The Chain Rule

The Chain Rule

The chain rule allows for the differentiation of functions of functions. Consider the expression $f(g(x))$ . If we differentiate this, we get $$ \frac{d}{dx}\left(f(g(x))\right) = \frac{d}{dx}\left(g(x)\right)\times \frac{d}{dg}\left(f(g)\right) $$

The proof of the chain rule can be derived using limits relatively easily. An outline is given below

$$\frac{d}{dx}\left(f(g(x))\right)= \lim_{x\rightarrow a}\frac{f(g(x)) - f(g(a))}{x-a} = \lim_{x\rightarrow a}\left(\frac{f(g(x)) - f(g(a))}{g(x)-g(a)}\cdot\frac{g(x) - g(a)}{x-a}\right) = \frac{df}{dg}\frac{dg}{dx}$$

There are some technical details to justify this proof that are omitted here.

The Chain Rule 1

The curve $C$ has equation $y = \sqrt{x^3 +1}$. Determine $\frac{dy}{dx}$
solution - press button to display
Let $g(x) = x^3 + 1$, let $f(g) = \sqrt{g}$. Then $$ \frac{dy}{dx} = \frac{df}{dg}\frac{dg}{dx} = \frac{1}{2}g^{-\frac{1}{2}}3x^2 = \frac{3x^2}{2\sqrt{x^3+1}} $$

The Chain Rule 2

The curve $C$ is defined by the equation $y = 4\sin(x^2)$. Determine $\frac{dy}{dx}$
solution - press button to display
Let $g = x^2$, let $f(g) = 4\sin(g)$ then $$ \frac{dy}{dx} = \frac{dg}{dx}\frac{df}{dg} = 2x\cdot 4\cos(g) = 8x\cos(x^2) $$