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

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- Dec 3rd 2009, 05:30 PMboomersooner1331more help needed
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 - Dec 3rd 2009, 05:44 PMadkinsjr
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 $\displaystyle x$.

$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

$\displaystyle \frac{dy}{dx}=4\frac{d}{du}ln(u)\frac{du}{dx}$

$\displaystyle =\frac{4}{u}(2x)=\frac{8x}{u}=\frac{8x}{x^2+9}$ - Dec 3rd 2009, 05:45 PMPulock2009you couldnot do this!?
y=ln(x^2+9)^4

=>y=4ln(x^2+9)

=>dy/dx=4.1/(x^2+9).2x(Wink) - Dec 3rd 2009, 05:52 PMboomersooner1331
so because it's a logarithm you don' multiply it by 4, and have 3 as an exponent?

- Dec 3rd 2009, 06:48 PMadkinsjr