# Finding the derivative of a trig function

i'm having trouble a bit of trouble with taking the derivatives and collating my results of trig functions in the form of $sin$ $3x$ for example.

The specific problem i'm stuck on is;

Find the derivative of $2\sin3t\cos4t$

I'll show what I have done, but i'm really looking for some explanation of a general case for this type of problem.

I let $u$ $=$ $2\sin3t$ and $v=\cos4t$

$du/dt=6\cos3t$

$dv/dt = -4\sin4t$

$d/dt[2\sin3t\cos4t] = u\cdot dv/dt +v\cdot du/dt$

$=-4\sin4t(2\sin3t)+6\cos3t(\cos4t)$

Firstly, are there any glaring mistakes in the above?

and secondly, my book says the answer should be $7\cos(7t) - \cos(t)$. I'm unsure of how to simplify my answer down to actually check it against the book.

I'd appreciate some general help on how I treat functions in this format.

Thank you!

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HINT:

Using $$2\sin B\cos A=\sin(A+B)-\sin(A-B)$$

$$2\sin3t\cos4t=\sin(4t+3t)-\sin(4t-3t)=\sin7t-\sin t$$

Other wise where you have left of use, $$2\sin A\sin B=\cos(A-B)-\cos(A+B)$$

and $$2\cos A\cos B=\cos(A-B)+\cos(A+B)$$

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Doh! I did not think to manipulate it before attempting to differentiate. Thanks for your help, it answered what would inevitably become my next question :) – Jacobadtr Jan 11 '14 at 14:36
@Jacobadtr, we should almost always try to simplify an expression before differentiation. Also, differentiation is of sum is easier than that of a product, right? – lab bhattacharjee Jan 11 '14 at 14:38
Absolutely it is – Jacobadtr Jan 11 '14 at 14:40