# Thread: Compostion of functions, Involving ln

1. ## Compostion of functions, Involving ln

The question is:
Differentiate: $\displaystyle y=ln(8x^2+2)^4$

Now, the answers in the back give:
$\displaystyle 32x/(4x^2+1)$
----as the derivative. I had a problem with this Question/Answer combo... I would think that the derivative it gives would be for the function
$\displaystyle y=ln[(8x^2+2)^4]$
.... I believe there is a difference, and I would like to know if I am correct.
Because I went about solving the question a different way, due to the way I see it written (I see it as $\displaystyle y=[ln(8x^2+2)]^4$ )

2. Originally Posted by mike_302
The question is:
Differentiate: $\displaystyle y=ln(8x^2+2)^4$

Now, the answers in the back give:
$\displaystyle 32x/(4x^2+1)$
----as the derivative. I had a problem with this Question/Answer combo... I would think that the derivative it gives would be for the function
$\displaystyle y=ln[(8x^2+2)^4]$
.... I believe there is a difference, and I would like to know if I am correct.
Because I went about solving the question a different way, due to the way I see it written (I see it as $\displaystyle y=[ln(8x^2+2)]^4$ )
$\displaystyle y = \ln(8x^2+2)^4$ ... only the argument $\displaystyle (8x^2+2)$ is raised to the 4th power

$\displaystyle y = [\ln(8x^2+2)]^4$ ... the entire log is raised to the 4th power.

$\displaystyle y = \ln(8x^2+2)^4$

$\displaystyle y = 4\ln(8x^2+2)$
$\displaystyle y' = 4 \cdot \frac{16x}{8x^2+2} = 4 \cdot \frac{8x}{4x^2+1} = \frac{32x}{4x^2+1}$

3. ahhhhhh, I forgot all about applying some log rules before doing the derivative. Good point!

Thanks!