### Differentiation

In this module, we have seen various ways of forming new functions. How to differentiate sums, differences, products, quotients, compositions and inverses is discussed in the module Introduction to differential calculus. We repeat them here for convenience.

###### Theorem

Let $$f,g$$ be differentiable functions. Then the derivative of $$f(x) + g(x)$$ is $$f'(x) + g'(x)$$, and the derivative of $$f(x) - g(x)$$ is $$f'(x) - g'(x)$$. That is,

$\dfrac{d}{dx} \bigl[f(x) + g(x)\bigr] = f'(x) + g'(x), \qquad \dfrac{d}{dx}\bigl[f(x) - g(x)\bigr] = f'(x) - g'(x).$
###### Theorem Product rule

Let $$f,g$$ be differentiable functions. Then the derivative of their product is

$\dfrac{d}{dx} \bigl[ f(x)\,g(x) \bigr] = f(x)\,g'(x) + g(x)\,f'(x).$
###### Theorem Quotient rule

Let $$f,g$$ be differentiable functions. Then the derivative of their quotient is

$\dfrac{d}{dx} \Bigl[ \dfrac{f(x)}{g(x)} \Bigr] = \dfrac{g(x)\,f'(x) - f(x)\,g'(x)}{\bigl(g(x)\bigr)^2}.$
###### Theorem Chain rule

Let $$f,g$$ be differentiable functions. Then the derivative of their composition is

$\dfrac{d}{dx} \bigl[ f\circ g(x)\bigr] = \dfrac{d}{dx} \bigl[ f(g(x)) \bigr] = f'(g(x))\,g'(x).$

#### Inverses

The derivative of an inverse function in terms of the original function is presented in the module Introduction to differential calculus through the use of the chain rule. We give a direct proof here and an example.

###### Theorem

Let $$y = f(x)$$ be a strictly increasing, differentiable function on the interval $$(a,b)$$. Then the inverse function $$x = g(y)$$ exists, and

$g'(y) = \dfrac{1}{f'(x)} = \dfrac{1}{f'(g(y))}.$
###### Proof

As $$f$$ is strictly increasing, the inverse function $$x = g(y)$$ exists. Consider

$\dfrac{g(y+k)-g(y)}{k}.$

Define

$h = g(y+k)-g(y).$

We are assuming $$f$$ is differentiable, and so it is continuous. Thus its inverse $$g$$ is continuous. This implies that, as $$k\to 0$$, $$g(y+k) \to g(y)$$ and so $$h \to 0$$.

Since $$g(y + k) = x + h$$, we obtain $$y + k = f(x+h)$$ and therefore $$k = f(x+h) - f(x)$$. We now have

$\dfrac{g(y+k)-g(y)}{k} = \dfrac{h}{f(x+h)-f(x)},$

and so it follows that

$g'(y) = \dfrac{1}{f'(x)} = \dfrac{1}{f'(g(y))}.$

$$\Box$$

The result also holds if the function $$f$$ is strictly decreasing.

#### Example

Let $$f \colon (-\dfrac{\pi}{2},\dfrac{\pi}{2}) \to \mathbb{R}$$ be given by $$f(x) = \sin x$$. Find the derivative of $$f^{-1}$$.

(This result is also established in the module The calculus of trigonometric functions. Here we derive the result using the notation of the previous theorem.)

#### Solution

We first note that $$f$$ is a strictly increasing function on this interval, and therefore the inverse exists. Let $$y = f(x)$$ and let $$g$$ be the inverse function of $$f$$. Then $$x = g(y)$$. By the previous theorem, we have

\begin{align*} g'(y) &= \dfrac{1}{f'(x)} \\ &= \dfrac{1}{\cos x} \\ &= \dfrac{1}{\cos(g(y))} \\ &= \dfrac{1}{\cos(\sin^{-1} y)}. \end{align*}

Since $$\sin^{-1} y \in (-\dfrac{\pi}{2}, \dfrac{\pi}{2})$$, we must have $$\cos(\sin^{-1} y) > 0$$. So we can use the Pythagorean identity to obtain

\begin{align*} g'(y) &= \dfrac{1}{\sqrt{\cos^2(\sin^{-1} y)}} \\ &= \dfrac{1}{\sqrt{1 - \sin^2(\sin^{-1} y)}} \\ &= \dfrac{1}{\sqrt{1-y^2}}. \end{align*}

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