# 2.5: Proofs of the Arithmetic of Derivatives

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The theorems of the previous section are not too difficult to prove from the definition of the derivative (which we know) and the arithmetic of limits (which we also know). In this section we show how to construct these rules.

Throughout this section we will use our two functions $$f(x)$$ and $$g(x)\text{.}$$ Since the theorems we are going to prove all express derivatives of linear combinations, products and quotients in terms of $$f,g$$ and their derivatives, it is helpful to recall the definitions of the derivatives of $$f$$ and $$g\text{:}$$

\begin{align*} f'(x) &=\lim_{h\to0} \frac{f(x+h)-f(x)}{h} &\text{and}&& g'(x) &=\lim_{h\to0} \frac{g(x+h)-g(x)}{h}. \end{align*}

Our proofs, roughly speaking, involve doing algebraic manipulations to uncover the expressions that look like the above.

## Proof of the Linearity of Differentiation (Theorem 2.4.2)

Recall that in Theorem 2.4.2 we defined $$S(x)=\alpha\,f(x)+\beta\,g(x)\text{,}$$ where $$\alpha,\beta\in\mathbb{R}$$ are constants. We wish to compute $$S'(x)\text{,}$$ so we start with the definition:

\begin{align*} S'(x) &= \lim_{h \to 0} \frac{S(x+h)-S(x)}{h} \end{align*}

Let us concentrate on the numerator of the expression inside the limit and then come back to the full limit in a moment. Substitute in the definition of $$S(x)\text{:}$$

\begin{align*} S(x+h)-S(x) &= \big[ \alpha f(x+h) + \beta g(x+h) \big] - \big[ \alpha f(x) + \beta g(x) \big] &\text{collect terms}\\ &=\alpha\big[f(x+h)-f(x)]+\beta\big[g(x+h)-g(x)\big] \end{align*}

Now it is easy to see the structures we need — namely, we almost have the expressions for the derivatives $$f'(x)$$ and $$g'(x)\text{.}$$ Indeed, all we need to do is divide by $$h$$ and take the limit. So let's finish things off.

\begin{align*} S'(x) &= \lim_{h \to 0} \frac{S(x+h)-S(x)}{h} & \text{from above}\\ &= \lim_{h \to 0} \frac{\alpha\big[f(x+h)-f(x)]+\beta\big[g(x+h)-g(x)\big] }{h}\\ &= \lim_{h \to 0} \left[ \alpha\frac{f(x+h)-f(x)}{h} + \beta\frac{g(x+h)-g(x)}{h} \right] &\text{limit laws}\\ &= \alpha\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} + \beta\lim_{h\to0}\frac{g(x+h)-g(x)}{h}\\ &=\alpha f'(x) + \beta g'(x) \end{align*}

as required.

## Proof of the Product Rule (Theorem 2.4.3)

After the warm-up above, we will just jump straight in. Let $$P(x)=f(x)\, g(x)\text{,}$$ the product of our two functions. The derivative of the product is given by

\begin{align*} P'(x) &= \lim_{h \to 0} \frac{P(x+h)-P(x)}{h} \end{align*}

Again we will focus on the numerator inside the limit and massage it into the form we need. To simplify these manipulations, define

\begin{align*} F(h) &= \frac{f(x+h)-f(x)}{h} & \text{ and } && G(h) &= \frac{g(x+h)-g(x)}{h}.\\ \end{align*}

Then we can write

\begin{align*} f(x+h)&= f(x)+hF(h) &\text{and} && g(x+h)&=g(x)+hG(h).\\ \end{align*}

We can also write

\begin{align*} f'(x) &= \lim_{h\to0} F(h) & \text{and}&& g'(x) &= \lim_{h\to0} G(h). \end{align*}

So back to that numerator:

\begin{align*} & P(x+h)-P(x) =f(x+h)\cdot g(x+h)-f(x) \cdot g(x) & \text{substitute}\\ &=[f(x)+ hF(h)]\ [g(x)+hG(h)]-f(x) \cdot g(x) & \text{expand}\\ &=f(x)g(x) + f(x)\cdot hG(h) + hF(h)\cdot g(x) + h^2 F(h)\cdot G(h) -f(x) \cdot g(x)\hskip-0.5in\\ &= f(x) \cdot hG(h) + hF(h) \cdot g(x) + h^2F(h) \cdot G(h). \end{align*}

\begin{align*} P'(x) &=\lim_{h\to 0}\frac{P(x+h)-P(x)}{h}\\ &= \lim_{h\to 0} \frac{f(x) \cdot hG(h) + hF(h) \cdot g(x) + h^2 F(h) \cdot G(h) }{h}\\ &= \left(\lim_{h\to 0} \frac{f(x) \cdot h G(h)}{h}\right) + \left(\lim_{h\to 0} \frac{h F(h) \cdot g(x)}{h}\right) + \left(\lim_{h\to 0} \frac{h^2 F(h) \cdot G(h) }{h}\right)\\ &= \left(\lim_{h\to 0} f(x) \cdot G(h)\right) + \left(\lim_{h\to 0} F(h) \cdot g(x)\right) + \left(\lim_{h\to 0} h F(h) \cdot G(h)\right)\\ \end{align*}

Now since $$f(x)$$ and $$g(x)$$ do not change as we send $$h$$ to zero, we can pull them outside. We can also write the third term as the product of 3 limits:

\begin{align*} &= \left(f(x) \lim_{h\to 0} G(h)\right) + \left(g(x) \lim_{h\to 0} F(h)\right) + \left(\lim_{h\to 0} h\right) \cdot \left(\lim_{h\to0} F(h)\right) \cdot \left(\lim_{h\to0} G(h)\right)\\ &= f(x) \cdot g'(x) + g(x)\cdot f'(x) + 0 \cdot f'(x) \cdot g'(x)\\ &= f(x) \cdot g'(x) + g(x) \cdot f'(x). \end{align*}

And so we recover the product rule.

## (Optional) — Proof of the Quotient Rule (Theorem 2.4.5)

We now give the proof of the quotient rule in two steps 1. We assume throughout that $$g(x) \neq 0$$ and that $$f(x)$$ and $$g(x)$$ are differentiable, meaning that the limits defining $$f'(x)\text{,}$$ $$g'(x)$$ exist.

• In the first step, we prove the quotient rule under the assumption that $$f(x)/g(x)$$ is differentiable.
• In the second step, we prove that $$1/g(x)$$ differentiable. Once we know that $$1/g(x)$$ is differentiable, the product rule implies that $$f(x)/g(x)$$ is differentiable.

Step 1: the proof of the quotient rule assumng that $$\frac{f(x)}{g(x)}$$ is differentiable.$$\ \ \$$ Write $$Q(x)=\frac{f(x)}{g(x)}\text{.}$$ Then $$f(x) = g(x)\,Q(x)$$ so that $$f'(x) = g'(x)\,Q(x) + g(x)\,Q'(x)\text{,}$$ by the product rule, and

\begin{align*} Q'(x) &= \frac{f'(x)-g'(x)\,Q(x)}{g(x)} = \frac{f'(x)-g'(x)\,\frac{f(x)}{g(x)}}{g(x)}\\ &= \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2} \end{align*}

Step 2: the proof that $$1/g(x)$$ is differentiable.$$\ \ \$$ By definition

\begin{align*} \frac{\mathrm{d} }{\mathrm{d} x}\frac{1}{g(x)} &=\lim_{h\rightarrow 0}\frac{1}{h}\left[\frac{1}{g(x+h)}-\frac{1}{g(x)}\right] =\lim_{h\rightarrow 0}\frac{g(x)-g(x+h)}{h\,g(x)\,g(x+h)}\\ &=-\lim_{h\rightarrow 0}\frac{1}{g(x)}\ \frac{1}{g(x+h)}\ \frac{g(x+h)-g(x)}{h}\\ &=-\frac{1}{g(x)}\ \lim_{h\rightarrow 0}\frac{1}{g(x+h)}\ \lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}{h}\\ &=-\frac{1}{g(x)^2}g'(x) \end{align*}

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