1. ## more 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

2. Originally Posted by boomersooner1331
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:

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

I always think of the inner function as a variable $u$ so that the application of the chain rule is easier. So think of the function as $y=4ln(u)$, where $u$ is a function of $x$.

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

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

$=\frac{4}{u}(2x)=\frac{8x}{u}=\frac{8x}{x^2+9}$

3. ## you couldnot do this!?

y=ln(x^2+9)^4
=>y=4ln(x^2+9)
=>dy/dx=4.1/(x^2+9).2x

4. so because it's a logarithm you don' multiply it by 4, and have 3 as an exponent?

5. Originally Posted by boomersooner1331
so because it's a logarithm you don' multiply it by 4, and have 3 as an exponent?
One of the properties of logarithms is that $ln(x)^a=aln(x)$. This is just an expression usually introduced in pre-calculus courses. I think that answers your question .