The quotient rule

Three quick questions from earlier lessons.

$\dfrac{x^3}{x} = x^2$ , whose derivative is $2x$ . The quotient of the derivatives would be $\dfrac{3x^2}{1}$ . Do they agree?

Top's derivative first, then subtract.

Quotient rule:

\left(\dfrac{f}{g}\right)' = \dfrac{f'\,g - f\,g'}{g^2}

, where

g \ne 0

$h(x) = \dfrac{x^2}{x + 1}$ . Find $h'(x)$ .

f' = 2x

g' = 1

f = x², g = x + 1.

h'(x) = \dfrac{2x(x + 1) - x^2}{(x + 1)^2}

f'g − fg' over g².

h'(x) = \dfrac{x^2 + 2x}{(x + 1)^2}

Simplify the top.

Your turn: $h(x) = \dfrac{e^x}{x}$ .

h'(x) = \dfrac{xe^x - e^x}{x^2}

f = eˣ, g = x: f'g − fg' = xeˣ − eˣ.

$f(1) = 1$ , $f'(1) = -2$ , $g(1) = 4$ and $g'(1) = 2$ . Find $\left(\dfrac{f}{g}\right)'(1)$ .
$f(1) = 1$ , $f'(1) = 1$ , $g(1) = 1$ and $g'(1) = 4$ . Find $\left(\dfrac{f}{g}\right)'(1)$ .

Three questions to check you have got it.

Quotient rule

$\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$ : "low d-high minus high d-low, over low squared".

Watch out: the order in the numerator matters.

End of the lesson