The Chain Rule: Mth 251 Homework

[Morse]

Hello! I'm currently taking Mth 251 and have been working on this chain rule for a bit. It's kind of straight forward, there's just a lot of chains inside of chains and so forth. I think I have a solution, but I'm not entirely sure its correct.

Homework Statement

Find the derivative of the function y=[x+(x+(sinx)^2)^3]^4

Homework Equations

(d/dx)(u^n) = nu^(n-1)*(du/dx)

The derivative of u to the nth power equals n*u to the n-1 power, times the derivative with respect to u.

The Attempt at a Solution

Okay so we have two parts for the chain rule, yes? The power rule part and the chain. So to start we have...

4[x+(x+(sinx)^2)^3]^3 = Power Rule Section
Chained to the derivative of the inside
1+3(x+(sinx)^2)^2
Now the chain after this is where I am unsure, but I believe it's...
(1+2sin(x)cos(x))
Do I need a new chain for the trig portion?


Anyway, putting that all together we should have
4[x+(x+(sinx)^2)^3]^3 * 1+3(x+(sinx)^2)^2 * (1+2sin(x)cos(x))

How terribly wrong is this?

 

[SammyS]

Toss in a missing pair of parentheses, & you're looking good !

4[x+(x+(sinx)^2)^3]^3 * (1+3(x+(sinx)^2)^2 )* (1+2sin(x)cos(x))