Hi,

I've been asked to differentiate $\displaystyle y = 5x^2tan(6x)$ logarithmically.

I took the following steps:

$\displaystyle y = 5x^2tan(6x)$

$\displaystyle ln(y) = ln(5x^2tan6x) = (2ln5x)(lntan(6x))$

$\displaystyle f(x) = 2ln5x$

$\displaystyle g(x) = lntan(6x)$

$\displaystyle f'(x) = \frac {2}{x}$

$\displaystyle g'(x) = \frac {1}{6Sec^26x}$

$\displaystyle \frac {1}{y} \frac{dy}{dx} = (2ln5x)(\frac{1}{6Sec^26x})+(\frac{2}{x})(lntan(6x ))$

$\displaystyle \frac {dy}{dx} = y((2ln5x)(\frac{1}{6Sec^26x})+(\frac{2}{x})(lntan( 6x)))$

$\displaystyle \frac{dy}{dx} = 5x^2tan(6x) ((2ln5x)(\frac{1}{6Sec^26x})+(\frac{2}{x})(lntan(6 x)))$

This is my first attempt at differentiating binomial products logarithmically, can anyone please look at these steps and tell me if and how I've gone wrong?

Thanks.

(Please spare any factorising, I don't understand it very well right now)