# Quotient Rule

Quotient Rule. If $f$ and $g$ are differentiable then $((f(x))/(g(x)))'=(f'(x)g(x)-f(x)g'(x))/(g(x))^2$.

Proof.

Let $h(x)=(f(x))/(g(x))$ then $f(x)=h(x)g(x)$.

Differentiating both sides gives: $f'(x)=(h(x)g(x))'$ or using product rule $f'(x)=h(x)g'(x)+h'(x)g(x)$.

From this we have that $h'(x)=(f'(x)-h(x)g'(x))/(g(x))$.

Remembering that $h(x)=(f(x))/(g(x))$ we can rewrite last identity as $((f(x))/(g(x)))'=(f'(x)-(f(x))/(g(x))g'(x))/(g(x))$ or $((f(x))/(g(x)))'=(f'(x)g(x)-f(x)g'(x))/(g(x))^2$.

Example 1. Find derivative of $f(x)=(e^x)/(x^3+x)$ .

$f'(x)=((e^x)/(x^3+x))'=((e^x)'(x^3+x)-e^x(x^3+x)')/(x^3+x)^2=(e^x(x^3+x)-e^x(3x^2+1))/(x^3+x)^2=$

$=(e^x(x^3-3x^2+x-1))/(x^3+x)^2$.

Example 2. Find derivative of $f(t)=(t^5+1)/t^3$

$f'(t)=((t^5+1)/(t^3))'=((t^5+1)'t^3-(t^5+1)(t^3)')/(t^3)^2=(5t^4t^3-(t^5+1)*3t^2)/t^6=$

$=(5t^7-3t^7-3t^2)/(t^6)=(2t^7-3t^2)/(t^6)=2t-3/t^4$

Note, however that Example 2 can be used without quotient rule. Since $(t^5+1)/t^3=t^2+1/t^3=t^2+t^(-3)$ then

$f'(t)=2t^(2-1)-3t^(-3-1)=2t-3t^(-4)=2t-3/t^4$ .

Example 2 showed that first you need to look at the function and make sure that quotient rule is the only way to find derivative (like in Example 1).

General advice is not to use quotient rule every time you see quotient. It is rather hard technique and if there are other ways to find derivative, use them.