find dy/dx of
y= ln(x^2 + 9)^4
i know you have to use the chain rule but i really need some help. i've got an exam tomorrow and i don't konw how to do this
Remember the properties of logarithms:
$\displaystyle y=ln(x^2+9)^4=4ln(x^2+9)$
I always think of the inner function as a variable $\displaystyle u$ so that the application of the chain rule is easier. So think of the function as $\displaystyle y=4ln(u)$, where $\displaystyle u$ is a function of $\displaystyle x$.
$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$
$\displaystyle \frac{dy}{dx}=4\frac{d}{du}ln(u)\frac{du}{dx}$
$\displaystyle =\frac{4}{u}(2x)=\frac{8x}{u}=\frac{8x}{x^2+9}$