## Calculus Help

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:

dx

But what about the composite of more than two functions? How would one differentiate

(sin(x^2+1))^3 ?

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.

dx

or more compactly,

dx

The notation of Leibniz can make it even clearer.

dx dg dh dx

This can be extended indefinitely, assuming the domains of our functions allow it:

dx

Now you can differentiate arbitrarily complex functions. Try it for

tan(sin(x^2+1))^3)

You should get

sec^2[((sin(x^2+1))^3] 3sin(x^2+1)^2 cos(x^2+1) 2x

We know the chain rule for a composite of two functions:

__d(fg)__= f'(g(x))g'(x)dx

But what about the composite of more than two functions? How would one differentiate

(sin(x^2+1))^3 ?

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)dx

or more compactly,

__d(fgh)__= f'gh g'h h'dx

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'dx

Now you can differentiate arbitrarily complex functions. Try it for

tan(sin(x^2+1))^3)

You should get

sec^2[((sin(x^2+1))^3] 3sin(x^2+1)^2 cos(x^2+1) 2x