Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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

share|improve this answer
    
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 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 at 14:38
    
Absolutely it is –  Jacobadtr Jan 11 at 14:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.