Chain Rule W/ Composite Functions

[Michele Nunes]

Homework Statement

If d/dx(f(x)) = g(x) and d/dx(g(x)) = f(x²), then d²/dx²(f(x³)) =
a) f(x⁶)
b) g(x³)
c) 3x²*g(x³)
d) 9x⁴*f(x⁶) + 6x*g(x³)
e) f(x⁶) + g(x³)

Homework Equations

The Attempt at a Solution

The answer is D. Since d/dx(f(x)) = g(x), I said that d/dx(f(x³)) should equal 3x²*g(x³), then I took the derivative again and first used product rule so 6x*g(x³) + 3x²*f(x⁶)*6x⁵ since you would need to do chain rule again but that doesn't match up with the answer. I've been trying to play around with it for a while and it's just not coming out as the answer.

 

[Samy_A]

Homework Statement

If d/dx(f(x)) = g(x) and d/dx(g(x)) = f(x²), then d²/dx²(f(x³)) =
a) f(x⁶)
b) g(x³)
c) 3x²*g(x³)
d) 9x⁴*f(x⁶) + 6x*g(x³)
e) f(x⁶) + g(x³)

Homework Equations

The Attempt at a Solution

The answer is D. Since d/dx(f(x)) = g(x), I said that d/dx(f(x³)) should equal 3x²*g(x³), then I took the derivative again and first used product rule so 6x*g(x³) + 3x²*f(x⁶)*6x⁵ since you would need to do chain rule again but that doesn't match up with the answer. I've been trying to play around with it for a while and it's just not coming out as the answer.

Redo the second derivative. The term 3x²*f(x⁶)*6x⁵ looks wrong.
By the product rule: $\frac{d}{dx} 3x²g(x³)= 6xg(x³) + 3x² \frac{d}{dx} g(x³)$

 

[Michele Nunes]

Redo the second derivative. The term 3x²*f(x⁶)*6x⁵ looks wrong.

Here's how I did the second derivative:
I'm taking the derivative of (3x²)*(g(x³)), and I'm going to let (3x²) = u and (g(x³)) = v
so the second derivative should look like: d²/dx²(f(x³)) = u'v + uv'
My first term came out nice: u'v = 6x*g(x³)
Now for my second term, I said that v' = f(x⁶)*6x⁵ since d/dx(g(x)) = f(x²) so I assumed that d/dx(g(x³)) would equal f((x²)³) or similarly, f(x⁶) and then I used chain rule for f(x⁶) which is why I added the 6x⁵ at the end

 

[Samy_A]

Now for my second term, I said that v' = f(x⁶)*6x⁵ since d/dx(g(x)) = f(x²) so I assumed that d/dx(g(x³)) would equal f((x²)³) or similarly, f(x⁶) and then I used chain rule for f(x⁶) which is why I added the 6x⁵ at the end

$\frac{d}{dx}g(x³) \neq g'(x³)$.
You are making this too complicated. Try to apply the chain rule when computing $3x² \frac{d}{dx} g(x³)$.

 

[Michele Nunes]

$\frac{d}{dx}g(x³) \neq g'(x³)$.
You are making this too complicated. Try to apply the chain rule when computing $3x² \frac{d}{dx} g(x³)$.

Ohhh I see now. I did chain rule on the wrong term. That's why it was coming out funky. Thank you!