I’m given a question that uses both the chain rule and product rule. I manged to do the chain rule part which was ok. But when it got to the product rule things got more serious. I subbed in the relevant terms, expanded them and now I split them up. I attempted to cancel them down but when I look at the answer my right-hand side was in correct. I put $2\sin2x\cdot e^x\sin x$ but that was unfortunately incorrect. So, I’m asking for your help to tell me how to completely cancel this down.
Thanks
~Neamus
For the derivative of $$\frac {e^x \sin x}{\cos 2x}$$ I got for the whole derivative (using the chain rule and product rule) $$\frac {\cos 2x (e^x (\sin x \cos x) + \sin x (e^x)) - e^x \sin x (-2 \sin 2x) }{\cos ^2 2x}$$
This equals $$\frac {\cos 2x (e^x (\sin x \cos x) + \sin x (e^x))}{\cos ^2 2x} - \frac {e^x \sin x (-2 \sin 2x)}{\cos^2 2x}$$
For the left hand side: $$\frac {\cos 2x (e^x (\sin x \cos x) + \sin x (e^x)}{\cos ^2 2x} \rightarrow \frac{e^x (\sin x \cos x) + \sin x (e^x)}{\cos 2x}$$
For the right hand side: $$\frac {e^x \sin x (-2 \sin 2x)}{\cos^2 2x} \rightarrow \frac {-2 e^x \sin x \sin 2x}{\cos ^2 2x} \rightarrow \frac {-2 e^x \sin x \tan 2x}{\cos 2x}$$
Combining both together and factoring...
$$\frac {e^x(\sin x \cos x) + e^x \sin x}{\cos 2x} + \frac {2e^x \sin x \tan 2x}{cos 2x} \rightarrow {\frac {e^x \sin x (\cos x + 1)+2 \tan 2x}{\cos 2x}} $$
EDIT: This can be simplified further to
$$\tan 2x (\frac {e^x}{2} + 2 \sec 2x) + e^x \sin x \sec 2x$$
If this needs further simplification or if I'm incorrect anywhere, let me know.