What is the derivative of y = sin^3 x tan 4x?

Question

I am suppose to use the chain rule. I can't quite get the answer on my practice AP calculus book. Can someone help me out step by step?

Answer suppose to be:
4sin^3(x) sec^2(4x) + 3sin^2(x)cos(x)tan(4x)

Answer 1

What is the derivative of y = sin³x (tan 4x)?

Use the product rule and the chain rule.

dy/dx = d[sin³x (tan 4x] / dx

= (3sin²x)(cos x)(tan 4x) + (sin³x)(4sec²4x)

= 3(sin²x)(cos x)(tan 4x) + 4(sin³x)(sec²4x)

Answer 2

the first*derivative of the second+ the second times the derivative of the first

sin^3x * sec^2 4x * 4+ tan 4x * 3 sin^2 x * cos x

=4sin^3 x * sec^2 4x + 3sin^2 x * cos x * tan 4x

you could simplify further to combine everything but it doesn't really accomplish anything(make it any neater)

Answer 3

The chain rule says

df(g(x))/dx
= df(g(x))/dg(x) * dg(x)/dx

In this case:

dtan(4x)
= dtan(4x)/d4x * d4x/dx
= sec^2 4x * 4

The idea is that you take the derivative of f with respect to g, then the derivative of g with respect to x. Each of those is simpler than the original df(g(x)). Note the "d4x" instead of "dx".

The rest is a lot of application of the product rule. Quite messy but if you're careful it will all come out.

Answer 4

The derivative of sinx is cosx
The derivative of tanx is (secx)^2

The derivative of (sinx)^3 = (3(sinx)^2)(cosx)
The derivative of tan4x = [(sec4x)^2](4)

Product rule is (uv)' = vu' + uv'

[((sinx)^3)*(tan4x)]' =
(tan4x)3((sinx)^2)cosx + ((sinx)^3)[(sec4x)^2](4)