Find the derivative.

I was able to figure out most of this on my own. But I got stuck at the part that involved cotangent.
Where did cotangent come from?? Please explain. Thanks guys :)

CurlyFry  May 10, 2018

Hi Curly :)



\(y=(sin9x)^{6x}\\ y=e^{ln(sin9x)^{6x}}\\ y=e^{6x\cdot ln(sin9x)}\\ let\;u=6x\cdot ln(sin9x)\\ \frac{du}{dx}=6\cdot ln(sin9x)+6x \cdot \frac{9cos9x}{sin9x}\\ \qquad \text{I used product rule here}\\ \qquad \text{and also used the fact that }\\ \qquad \frac{d}{dx}ln(f(x))= \frac{f'(x)}{f(x)}\\ \frac{du}{dx}=6\cdot ln(sin9x)+6x \cdot \frac{9cos9x}{sin9x}\\ \frac{du}{dx}=6\cdot ln(sin9x)+6x \cdot 9(cot9x)\\ \frac{du}{dx}=6\left[ ln(sin9x)+9x(cot9x)\right] \\ \)


\(y=e^{6x\cdot ln(sin9x)}\\ y=e^{u}\\ \frac{dy}{du}=e^u\\ \frac{dy}{du}=e^{6x\cdot ln(sin9x)}\\ \frac{dy}{du}=e^{ ln(sin9x)\cdot 6x}\\ \frac{dy}{du}=(sin9x)^{6x}\\ \)


\(\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}\\ \frac{dy}{dx}=(sin9x)^{6x}\cdot 6\left[ ln(sin9x)+9x(cot9x) \right]\\ \frac{dy}{dx}=6(sin9x)^{6x}\left[ ln(sin9x)+9x(cot9x) \right]\\\)


If you do not understand one of the steps please explain which step/s is giving you trouble.

Melody  May 10, 2018

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