Find the derivatives of:

Question

1. H(x)= sin2xcos2x

The answer given is 2cos4x. My question is, how in the world did they get that!? Shouldn't the answer at least contain the sin function, either negative or positive seeing as it's the derivative of cos? Also, which rules are involved in this problem? Product and chain rule?

2. f(x)= (cotx) / (sinx)
I ended up with a mess of an answer that no where near resembled the solution.

Any help would be greatly appreciated! :)

Answer 1

H(x) = sin2x cos2x
= 1/2 sin4x
That help?

Using the product rule, it's a bit more work, but you can get the same answer:

dH/dx = 2cos2x cos2x - 2sin2x sin2x
= 2(cos^2 2x - sin^2 2x)
= 2cos4x

Don't forget your trig just because you're taking calculus!

#2
df/dx = (u'v - uv')/v^2
= (-csc^2 * sin - cot * cos)/sin^2
= (-csc - cos^2/sin)/sin^2
= -1/sin^3 - cot^2 csc
= -csc(csc^2+cot^2)

or, getting rid of the pesky cot,

f = cos/sin^2
f' = (-sin^3 - 2cos^2*sin)/sin^4
= (-sin^2 - 2cos^2)/sin^3
= -(1+cos^2)/sin^3
= -csc(csc^2+cot^2)