Thread: [SOLVED] Can't figure out this differentiation/chain rule problem:

1. [SOLVED] Can't figure out this differentiation/chain rule problem:

I've been taking lessons out of this Calc book I picked up (Thomas-Finney 9th Edition), and I've been doing decently. But I've come across these sets of problems that I just can't for the life of me figure out, since the book doesn't give any examples similar to them.

If anyone could help, I'd be very, very, very grateful. If possible, a detailed explanation would be great!

Find the derivative of the function:

y = x^2sin^4x + xcos^-2x (Note: It's the sine of x to the fourth power, just for clarification. Same with the cosine section.)

and..

y = 4/3[pi] sin3t + 4/5[pi] cos5t

By [pi] I mean 3.14...

2. First part of the first one: use product rule: d(uv)/dx = u dv/dx + v du/dx.

Put u = x^2 and v = sin^4 x

How to do sin^4 x: put z = sin x and use the chain rule: dy/dx = (dy/dz) (dz/dx).

The rest of the examples are further applications of the chain and product rule. The book should explain it fully.

If you're wondering about the product rule, see where it's proved from first principles in the book (which I don't know but if it's any good it *must* have that in it).

3. That's what I've been trying to do, but the response I keep on getting isn't consistent with the one from the back of the book.

How I've been doing it:

y = x^2 sin^4 x + x cos^2 x (Basic problem)

I split the problem in two two parts, according to the rule that the derivative of (x+y) is x'+y':

First doing
y = x^2 sin^4 x

Applying the product rule of derivation:

y' = ((x^2)(sin^4 x)') + ((sin^4 x)(x^2)')
Product rule = d(uv)/dx = (u)dv/dx + (v)du/dx
Let u = x^2 Let v = sin^4 x

y' = (x^2((4sin^3 x)(cos x)) + ((sin^4 x)(2x))
Apply the chain rule to the first sin^4 x,
Chain rule: If y = f(g(x)), dy/dx = f'(g(x)) g'(x), with the added Inside-Out rule that the g(x) of the function f(g(x)) can be left untouched until later.
f(u) = u^4, g(x) = (sin x)

y' = (4x^2 sin^3 x cos x) + (2x sin^4 x)
Multiply together the requisite terms.

The second part =
y = x cos^2 x
y' = ((x)(2cos x)(-sinx)) + ((cos^2 x)(1))
y' = (-2xcos x sin x) + (cox^2 x)

Combining them:

y' = (4x^2sin^3 x cos x) + (2x sin^4 x) - (2xcos x sin x) + (cos^2 x)

I can't figure out what I did wrong.

The answer given was:

y' = 2x sin^4 + 4x^2 sin^3 x cos x + cos^-2 x + 2x cos^-3 x sin x

4. first part's okay, 2nd part you've been given x cos^-2 x but you're deriving x cos^2 x. RTQ.

5. D'oh! Thanks a lot. To those that come after me: make sure you write down the correct problem, kids!