# differentiate sinusoidal function

• Apr 2nd 2013, 10:11 AM
onehundredpercenteffort
differentiate sinusoidal function
Question: Differentiate with respect to x

a. y=3sin^2(2x-4) - 2cos^2 (3x+1)

b. y=sin(x^2 + pi)

My teacher says for these questions, i need to use the product rule/chain rule/power rule/sum or difference rule/quotient rule etc. I know the derivative of y=sinx is y'= cosx and the derivative of y=cosx is y'=-sinx.

• Apr 2nd 2013, 10:45 AM
ssgohanf8
Re: differentiate sinusoidal function
The first term should differentiate into 3(2)(sin(2x-4))(cos(2x-4))(2) following the chain rule by first multiplying by the exponent of an argument and reducing the exponent by one and rewriting it. Then multiplying it by the derivative of sin(2x-4). Then multiplying it by the derivative of the argument in that one as well. 2. Following the same logic with the second term gives (2)(2 [the exponent of cos2(3x+1)])(cos(3x+1) [rewriting the argument that was squared, but the exponent is n-1])(-sin(3x+1) [The derivative of the argument inside that was squared])(3 [Derivative of 3x+1])

So, a. y`= 3(2)(sin(2x-4))(cos(2x-4))(2) -(2)(2)(cos(3x+1))(-sin(3x+1))(3)
= 12sin(2x-4)cos(2x-4) + 12cos(3x+1)sin(3x+1)
= 12(sin(2x-4)cos(2x-4) + sin(3x+1)cos(3x+1))

b. y`=cos(x2+π)(2x) = 2xcos(x2+π)
• Apr 2nd 2013, 02:07 PM
onehundredpercenteffort
Re: differentiate sinusoidal function
Thanks for replying. However, the answer to a. is different from the back of the book, which says the answer is y'=12sin(2x-4)cox(2x-4) + 12cos(3x-1)sin(3x-1)

Also, in b., why did you multiply by 2x?

Thus, I'm still a bit confused. Could youclarify further as to how they arrived to that answer? Thanks.
• Apr 2nd 2013, 03:17 PM
ssgohanf8
Re: differentiate sinusoidal function
The book stopped on my second to last line. On my last line, I simply factored out the 12 from both terms; sorry about that. The reasoning for multiplying by 2x in b., is because that is the derivative of (x2-pi) within the argument of your sin function.

When you take derivatives, think of things like sin(cos(x)) as just sin(x) and take its derivative normally. Then multiply it by what the derivative of what x would have been. In my example, it's cos(x)

y=sin(cos(x))
y`=cos(cos(x))(-sin(x))(1) The 1 is because the derivative of x1 is 1.

K=f(g(x))
K`=f`(g(x))g`(x) If that makes it any easier. Sorry for the late reply.
• Apr 2nd 2013, 03:26 PM
onehundredpercenteffort
Re: differentiate sinusoidal function
thanks