Yes, you are correct. Thank you for catching that. So,
d/dx(df/dg) = dg/dx * d2f/dg2.
I find it strange that in every calculus book I have ever read (including real analysis books), they always stop at the first derivative when discussing the chain rule.
Thank you, SammyS! I tried the sample function you suggested, but I'm still not sure if I am expressing d/dx (df/dg) correctly. Here is how your sample function worked out:
y(x) = sin(x5)
So, f(x) = sin(x) and g(x) = x5.
Therefore,
dy/dx = cos(x5) 5x4
and
d2y/dx2 = cos(x5) 20x3 + 5x4...
Homework Statement
Derive an expression for the composition of 2 functions.
Homework Equations
The Attempt at a Solution
I started with supposing that y(x) = f(g(x)). I know that dy/dx = df/dg * dg/dx (via the chain rule). Doing the derivative again, I started with the product...