What is the derivitive and method(of getting the derivitive) of
f'(x)=-10/(3x+2)^2
Thanks guys.
Initial equation
$\displaystyle f\prime (x)=-\frac{10}{(3x+2)^2}$
Take the derivative
$\displaystyle f\prime\prime(x)=\frac{d}{dx}\frac {-10}{(3x+2)^2}$
-10 is a coefficient, so we can write
$\displaystyle f\prime\prime(x)=-10\left[\frac{d}{dx}\frac 1{(3x+2)^2}\right]$
Then we rewrite without the fractions
$\displaystyle f\prime\prime(x)=-10\left[\frac{d}{dx}(3x+2)^{-2}\right]$
Use the product rule to move the negative 2 out front, and subtract 1 from the exponent making it negative 3, and the chain rule to take the derivative of the inside of the exponent.
$\displaystyle f\prime\prime(x)=-10(-2)(3x+2)^{-3}\left[\frac{d}{dx}(3x+2)\right]$
Use the power rule to turn d\dx(3x+2) into 3
$\displaystyle f\prime\prime(x)=-10(-2)(3x+2)^{-3}(3)$
Combine coefficients
$\displaystyle f\prime\prime(x)=60(3x+2)^{-3}$
Write as a fraction
$\displaystyle f\prime\prime(x)=\frac{60}{(3x+2)^3}$