# Thread: Differentiate y = cos(a^3 + x^3)

1. ## Differentiate y = cos(a^3 + x^3)

The problem is as follows: Differentiate y = cos(a^3 + x^3)

my current solution:

By the chain rule
u = a^3 + x^3
f(u) = cos(u)

u' = 3a^2 + 3x^2 (by the power rule)
f ' (u) = -sin(u)

therefore,

-sin(a^3 + x^3)(3a^2 + 3x^2) = -3a^2sin(a^3 + x^3) - 3x^2sin(a^3 + x^3)

This is the solution I come up with, however it is incorrect and I don't know why.
correct solution should be -3x^2sin(a^3 + x^3). Where is the -3a^2sin(a^3 + x^3)
going?

2. ## Re: Plz help with a particular differentiation

Originally Posted by iAmKrizzle
By the chain rule
u = a^3 + x^3
f(u) = cos(u)

u' = 3a^2 + 3x^2 (by the power rule)
This is not correct, i'm guessing you are only differentiating w.r.t x?

3. ## Re: Plz help with a particular differentiation

You are almost right. Your mistake comes from forgetting that $\displaystyle a^3$ is not a variable: it's a constant value, such as 8 or 27. Therefore when we differentiate it, we just get 0 - it vanishes!

$\displaystyle y=cos(a^3+x^3)$

Let $\displaystyle u=a^3+x^3$

$\displaystyle \frac{du}{dx}=3x^2$

$\displaystyle \frac{dy}{du}=-sin(u)$

$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

4. ## Re: Plz help with a particular differentiation

thank you Quacky that makes sense. I was treating a^3 as a variable. Thank you for your help.