# how did i do and what did i do wrong

#### jpratt

Let f(x) = -9cos(sin(x^3)), then f'(x)

so far i did this:
f'(x)=-9(cos(sin(x^3))
= -9(-sin)(sinx^3) *(sinx^3)
= -9(-sin)(sinx^3) * cos(x^3) * (x^3)'
= 9sin(sinx^3)*cos 3x^2
=27x^2sin(sinx^3)cosx^3
did i miss anything (Wondering)

#### wonderboy1953

Why did you repeat (sinx^3) in the second line?

#### slider142

Let f(x) = -9cos(sin(x^3)), then f'(x)

so far i did this:
f'(x)=-9(cos(sin(x^3))
This does not make sense, as it is not f'(x). Precise true statments are very important in mathematics. Strict professors would immediately take marks off for sloppy notation like this, as it may indicate incorrect logic behind the solution.
= -9(-sin)(sinx^3) *(sinx^3)'
What is (-sin) ? What does it mean? Do you mean (-sin(x)) ? That doesn't make sense in the context of the problem either. I believe you meant -sin(sin(x^3)). Proper use of parentheses is important to avoid ambiguities.
= -9(-sin)(sinx^3) * cos(x^3) * (x^3)'
= 9sin(sinx^3)*cos 3x^2
I hope this was a typo and is not how it's written on your page. Are you taking the cosine of 3x^2, or multiplying 3x^2 by cos(x)? or neither?
=27x^2sin(sinx^3)cosx^3
did i miss anything (Wondering)
The conclusion is correct, but hopefully the logic was just full of typos.

TheCoffeeMachine

#### jpratt

it was my way of breaking it down, so i can understand it..