how did i do and what did i do wrong

Jun 2010
21
0
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)
 
May 2009
72
21
Brooklyn, NY
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.
 
  • Like
Reactions: TheCoffeeMachine
Jun 2010
21
0
it was my way of breaking it down, so i can understand it..