# more help needed

• Dec 3rd 2009, 05:30 PM
boomersooner1331
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
• Dec 3rd 2009, 05:44 PM
Quote:

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:

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

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 PM
Pulock2009
you 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 PM
boomersooner1331
so because it's a logarithm you don' multiply it by 4, and have 3 as an exponent?
• Dec 3rd 2009, 06:48 PM
One of the properties of logarithms is that $\displaystyle ln(x)^a=aln(x)$. This is just an expression usually introduced in pre-calculus courses. I think that answers your question (Hi).