We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

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 :)

 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.

 May 10, 2018

7 Online Users