f(x) = x^(3x)

So far i used implicit differentiation and get

(1/y)(y')=2x + 3ln(x)

but I am way off,

any help would be appreciated!

Printable View

- Oct 26th 2009, 07:36 PMfrozenflamesMaking some progress on differentiation problem
f(x) = x^(3x)

So far i used implicit differentiation and get

(1/y)(y')=2x + 3ln(x)

but I am way off,

any help would be appreciated! - Oct 26th 2009, 07:38 PMMush
- Oct 26th 2009, 07:39 PMrn443
- Oct 26th 2009, 07:42 PMfrozenflames
- Oct 26th 2009, 07:44 PMMush
Yes, but you know what y is from the original equation. Simply substitute it in. $\displaystyle y = x^{3x} $

So $\displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{1}{x^{3x}} \frac{dy}{dx} $

So multiply through by $\displaystyle x^{3x} $ to get your equation $\displaystyle dy/dx = $ purely in terms of x. - Oct 26th 2009, 08:08 PMfrozenflames
I submit the following and it shows up as incorrect..

I may be missing something?

x^(3x)[2x + 3ln(x)] - Oct 26th 2009, 08:13 PMMush
Yes. As I said before $\displaystyle 2x + 3 \ln(x) $ is NOT the derivative of $\displaystyle 3x. \ln {x} $

You have carried out your differentiation incorrectly on the RHS of the equation AFTER you've taken the natural logarithm of both sides.

$\displaystyle y = x^{3x} $

$\displaystyle \ln(y) = \ln(x^{3x}) $

$\displaystyle \ln(y) = 3x \ln (x) $

$\displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (3x . \ln(x)) $

You must use the product rule on the RHS to differentiate. - Oct 26th 2009, 08:42 PMfrozenflames
- Oct 26th 2009, 08:45 PMMush