The super chain rule AKA recursive chain rule for higher order compositions of functions.
We know the chain rule for a composite of two functions:
d(fg) = f'(g(x))g'(x)
But what about the composite of more than two functions? How would one differentiate
Here's a formula for repeated applications of the chain rule that I have never learned or seen formulated in any calculus classes anywhere. Not claiming that I'm the first person to discover it, but I certainly am not borrowing this from elsewhere.
d(fgh) = f'(g(h(x)))g'(h(x))h'(x)
or more compactly,
d(fgh) = f'gh g'h h'
The notation of Leibniz can make it even clearer.
df = df dg dh
dx dg dh dx
This can be extended indefinitely, assuming the domains of our functions allow it:
d(fghijklm) = f'ghijklm g'hijklm h'ijklm i'jklm j'klm k'lm l'm m'
Now you can differentiate arbitrarily complex functions. Try it for
You should get
sec^2[((sin(x^2+1))^3] 3sin(x^2+1)^2 cos(x^2+1) 2x