Chain rule help please

[alwaysthecalculus]

Wish I could post my own work, but I don't know where to start. UGH!

 

[Deleted member 4993]

Wish I could post my own work, but I don't know where to start. UGH!

Do you know - how to calculate $\displaystyle \frac{d}{dx}f[g(x)] \ = \ ?$

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem.

 

[pka]

$D_x\left(f\circ g(x)\right)=f'(g(x))g'(x)$

 

[alwaysthecalculus]

$D_x\left(f\circ g(x)\right)=f'(g(x))g'(x)$

Yes, I know outside/inside; what's throwing me off is second derivative of f(x^3) instead of f(x).

 

[pka]

Yes, I know outside/inside; what's throwing me off is second derivative of f(x^3) instead of f(x).

O.K. Let's look at $f(x)=\log(\cos^3(x^2+1))$ then $\Large f'(x)=\frac{3\cos^2(x^2+1)(-\sin(x^2+1)(2x))}{\cos^3(x^2+1)}$

 

[Steven G]

[MATH]\dfrac{d}{dx}f(x^3)=f'(x^3)*3x^2 = g(x^3)*3x^2[/MATH]. Now take the derivative again wrt x.

 

[alwaysthecalculus]

[MATH]\dfrac{d}{dx}f(x^3)=f'(x^3)*3x^2 = g(x^3)*3x^2[/MATH]. Now take the derivative again wrt x.

Thank you, this got me to the finish line: D!