Algebra of derivatives

Once I can differentiate one function I want sums, scalar multiples, and products without returning to (2.1) every time. The proofs are the definition plus algebra; the point is to see where the little-o remainder (2.3) is still small enough.

Theorem 2.1

𝑓

and

𝑔

are differentiable at

𝑎

, then so is

𝑓+𝑔

, and

{(𝑓+𝑔)}^{\prime }(𝑎)={𝑓}^{\prime }(𝑎)+{𝑔}^{\prime }(𝑎)

. For a constant

𝑐

{(𝑐𝑓)}^{\prime }(𝑎)=𝑐{𝑓}^{\prime }(𝑎)

Proof

The difference quotient of the sum is the sum of the difference quotients. Each side has a limit, so the sum does, and the limit of the sum is the sum of the limits. For

𝑐𝑓

, every term is multiplied by

𝑐

In particular, having already computed ${({𝑥}^{2})}^{\prime }=2𝑥$ and ${({𝑥}^{3})}^{\prime }=3{𝑥}^{2}$ , I get ${({𝑥}^{3}+4{𝑥}^{2})}^{\prime }=3{𝑥}^{2}+8𝑥$ by Theorem 2.1 without expanding ${(𝑎+h)}^{3}+4{(𝑎+h)}^{2}$ from scratch.

Product

The product is the first place the algebra has to do work. If I write $𝑓(𝑎+h)=𝑓(𝑎)+{𝑓}^{\prime }(𝑎)h+𝑟(h)$ and $𝑔(𝑎+h)=𝑔(𝑎)+{𝑔}^{\prime }(𝑎)h+𝑠(h)$ as in (2.3), the product is

𝑓(𝑎+h)𝑔(𝑎+h)=𝑓(𝑎)𝑔(𝑎)+({𝑓}^{\prime }(𝑎)𝑔(𝑎)+𝑓(𝑎){𝑔}^{\prime }(𝑎))h+\varepsilon (h),

where $\varepsilon (h)$ collects ${h}^{2}$ terms and the remainders $𝑟$ , $𝑠$ . Then $\frac{\varepsilon (h)}{h}\to 0$ , which is the criterion for differentiability.

Theorem 2.2

𝑓

and

𝑔

are differentiable at

𝑎

, then so is

𝑓𝑔

, and

{(𝑓𝑔)}^{\prime }(𝑎)={𝑓}^{\prime }(𝑎)𝑔(𝑎)+𝑓(𝑎){𝑔}^{\prime }(𝑎)

Proof

Split the difference:

𝑓(𝑎+h)𝑔(𝑎+h)-𝑓(𝑎)𝑔(𝑎)=𝑓(𝑎+h)(𝑔(𝑎+h)-𝑔(𝑎))+𝑔(𝑎)(𝑓(𝑎+h)-𝑓(𝑎)).

Divide by $h$ . The first piece tends to $𝑓(𝑎){𝑔}^{\prime }(𝑎)$ because $𝑓$ is continuous at $𝑎$ by Proposition 2.1 and $𝑔$ is differentiable at $𝑎$ . The second piece tends to $𝑔(𝑎){𝑓}^{\prime }(𝑎)$ .

Example 2.2

Let

𝑓(𝑥)=𝑥

and

𝑔(𝑥)=𝑥

. Then

(𝑓𝑔)(𝑥)={𝑥}^{2}

, and Theorem 2.2 says

{({𝑥}^{2})}^{\prime }=1\cdot 𝑥+𝑥\cdot 1=2𝑥

, which matches (2.2).

Quotient, once

If $𝑔(𝑎)\neq 0$ and $𝑔$ is differentiable at $𝑎$ , then $\frac{1}{𝑔}$ is differentiable at $𝑎$ and

{\left(\frac{1}{𝑔}\right)}^{\prime }(𝑎)=-{𝑔}^{\prime }\frac{𝑎}{{𝑔(𝑎)}^{2}}.

(2.4)

I believe this by writing $𝑔(𝑎+h)=𝑔(𝑎)+{𝑔}^{\prime }(𝑎)h+𝑠(h)$ and expanding $\frac{1}{𝑔(𝑎+h)}-\frac{1}{𝑔(𝑎)}$ over a common denominator; the $h$ in the numerator is again genuine. Combined with Theorem 2.2 this is the quotient rule. I will not need a separate statement until I start dividing polynomials.

Exercise 2.2

|𝑥|\cdot 𝑥

equals

𝑥|𝑥|

. Is this product differentiable at

0

Solution to Exercise 2.2

For

𝑥\geq 0

the product is

{𝑥}^{2}

; for

𝑥<0

it is

-{𝑥}^{2}

. The difference quotient at

0

\pm h

, which tends to

0

from both sides, so the derivative exists and is

0

. Theorem 2.2 does not apply, because

𝑥\mapsto |𝑥|

is not differentiable at

0

; the product can be smoother than a factor.