Calculus I problem I can't solve

[ben23]

Here's the problem:

d/dx(f(3x^5)) = 8x^2

Find f'(x)

After applying the chain rule:

f'(3x^5)(15x^4) = 8x^2

f'(3x^5) = 8/(15x^2)

It's not apparent to me how I proceed from here to find f'(x). I tried dividing the expression on the right by three and taking the fifth root but that does not seem to be right. Any help would be appreciated!

 

[HallsofIvy]

Since you want to know f(x) without all that "stuff" inside the parentheses, let u= 3x⁵. Then f(3x⁵)= f(u) and df/dx= (df/du)(du/dx)= (df/du)(15x)= 8x². Now you have df/du= (8/15)x. Since u= 3x⁵, x⁵= u/3 and x= (u/3)^1/5. df/du= (8/15)(u^1/5)/3^1/5. Now just replace u by x to get f'(x)

 

[ben23]

Since you want to know f(x) without all that "stuff" inside the parentheses, let u= 3x5. Then f(3x5)= f(u) and df/dx= (df/du)(du/dx)= (df/du)(15x)= 8x2. Now you have df/du= (8/15)x. Since u= 3x5, x5= u/3 and x= (u/3)1/5. df/du= (8/15)(u1/5)/31/5. Now just replace u by x to get f'(x)

I see what you're saying, but isn't du/dx equal to 15x^4, not 15x? That would make df/du equal to (8/15x^2). Therefore, substituting x= (u/3)1/5 gives a result of (8/15)(3^(2/5)/u^(2/5)).

Substituting u by x to get f'(x) would get a final solution of 8/15x^2, which is what I started with above for f'(3x^5), which I don't think is the correct answer.