Chain Rule Proof

• Lecture topic is the Chain Rule

Chain Rule Theorem

• If $g$ is differentiable at $x_0$ and $f$ is differentiable at $g(x_0)$, then $f \circ g$ is differentiable at $x_0$.
• $(f \circ g)'(x_0) = f'(g(x_0)) \cdot g'(x_0)$

Proof of Chain Rule

• Define $y_0 = g(x_0)$.
• Since $f$ is differentiable at $y_0$, $f(y) = f(y_0) + f'(y_0)(y - y_0) + \psi(y)(y - y_0)$, where $\psi(y) \to 0$ as $y \to y_0$.
• Substitute $g(x)$ for $y$: $f(g(x)) = f(g(x_0)) + f'(g(x_0))(g(x) - g(x_0)) + \psi(g(x))(g(x) - g(x_0))$.
• Since $g$ is differentiable at $x_0$, $g(x) = g(x_0) + g'(x_0)(x - x_0) + \phi(x)(x - x_0)$, where $\phi(x) \to 0$ as $x \to x_0$.
• Substitute this into the previous equation: $f(g(x)) = f(g(x_0)) + f'(g(x_0))[g'(x_0)(x - x_0) + \phi(x)(x - x_0)] + \psi(g(x))[g'(x_0)(x - x_0) + \phi(x)(x - x_0)]$.
• $f(g(x)) = f(g(x_0)) + f'(g(x_0))g'(x_0)(x - x_0) + \epsilon(x)(x - x_0)$.
• $\epsilon(x) = f'(g(x_0))\phi(x) + \psi(g(x))g'(x_0) + \psi(g(x))\phi(x)$.
• Since $g$ is continuous at $x_0$, as $x \to x_0$, $g(x) \to g(x_0)$, hence $\psi(g(x)) \to 0$. Also, $\phi(x) \to 0$ as $x \to x_0$. Thus $\epsilon(x) \to 0$ as $x \to x_0$.
• $f(g(x)) = f(g(x_0)) + f'(g(x_0))g'(x_0)(x - x_0) + \epsilon(x)(x - x_0)$, therefore $f \circ g$ is differentiable at $x_0$ and $(f \circ g)'(x_0) = f'(g(x_0)) \cdot g'(x_0)$.

Chain Rule Example 1

• Find $h'(x)$ where $h(x) = \sin(x^2 + 1)$
• Let $f(x) = \sin(x)$ and $g(x) = x^2 + 1$. Then $h(x) = f(g(x))$.
• $h'(x) = f'(g(x)) \cdot g'(x) = \cos(x^2 + 1) \cdot 2x = 2x\cos(x^2 + 1)$

Chain Rule Example 2

• Find $h'(x)$ where $h(x) = (\sin x + 1)^3$.
• Let $f(x) = x^3$ and $g(x) = \sin x + 1$. Then $h(x) = f(g(x))$.
• $h'(x) = f'(g(x)) \cdot g'(x) = 3(\sin x + 1)^2 \cdot \cos x = 3\cos x(\sin x + 1)^2$

Inverse Function Theorem

• Suppose $f$ is differentiable on an interval $I$ and $f'(x) \neq 0$ on $I$.
• Then $f$ is one-to-one, $f^{-1}$ is differentiable at $f(x)$, and $(f^{-1})'(f(x)) = \frac{1}{f'(x)}$.

Proof of Inverse Function Theorem

• Since $f'(x) \neq 0$ on $I$, either $f'(x) > 0$ on $I$ or $f'(x) < 0$ on $I$.
• If $f'(x) > 0$ on $I$, $f$ is increasing; if $f'(x) < 0$ on $I$, $f$ is decreasing. In either case, $f$ is one-to-one.
• Let $y = f(x)$, so $x = f^{-1}(y)$. Then $f(f^{-1}(y)) = y$.
• By The Chain Rule, $f'(f^{-1}(y)) \cdot (f^{-1})'(y) = 1$.
• Since $f'(x) \neq 0$ on $I$, $f'(f^{-1}(y)) \neq 0$, so $(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}$.
• Replacing $y$ with $f(x)$ yields $(f^{-1})'(f(x)) = \frac{1}{f'(x)}$.