Question

Find the first derivative.

f(x) = sin^2(4x^3)

Answer 1

\[\sin^{2}(4x^{3})\]

Answer 2

12x^3*2sin(4x^3)*cos(4x^3)

Answer 3

Don't forget the chain rule. \[f'(x)=2(\sin(4x^3))(\cos(4x^3))(12x^2)\]
Did I miss any?
The constituent functions are x^2 on the outside, sin(x) in the middle, and 4x^3 on the inside

Answer 4

\[f(g(x))=\sin^2(4x^3), f(x)=\sin^2x, g(x)=4x^3\]
\[f'(x)=2sinxcosx, g'(x)=12x^3, f'(g(x))=f'(4x^3)=2\sin(4x^3)\cos(4x^3)\]
\[f'(g(x))*g'(x)=2\sin(4x^3)\cos(4x^3)*12x^2\]

Answer 5

oops type g'(x)=12x^2

Answer 6

Should that be an x^2?

Answer 7

yes yes it should be
\[g'(x)=12x^2\]

Answer 8

Sorry, you beat me to it!

Answer 9

i haven't had any sleep i don't know what keys im hitting lol