What is the derivative of f(g(h(x)))?

1 Answer

It's #f^prime(g(h(x))) g^prime (h(x)) h^prime(x)#

Start by defining the function #a(x)=g(h(x))#

The the chain rule gives us:

#(f @ g @ h)^prime (x)=(f @ alpha)^prime (x)=f^prime(alpha(x)) alpha^prime(x)#

Applying the definition of #alpha(x)# to the equation above gives us:

#f^prime(alpha(x)) alpha^prime(x) = f^prime (g(h(x))) (g @h)^prime (x)#

Using the chain rule again:

#f^prime (g(h(x))) (g @h)^prime (x) = f^prime(g(h(x))) g^prime(h(x)) h^prime(x)#

Therefore:

#(f @ g @ h)^prime (x) = f^prime(g(h(x))) g^prime(h(x)) h^prime(x)#

The same derivation using the Leibniz notation and the definitions #y=h(x)#, #w=g(y)#:

#d/(dx)[f(w)]=(df)/(dw) (dw)/dx=(df)/(dw) (dw)/(dy) (dy)/(dx) = (df)/(dw) (w) (dw)/(dy) (y) (dy)/(dx) (x) = (df)/(dw) (g(y)) (dw)/(dy) (h(x)) (dy)/(dx) (x) = (df)/(dw) (g(h(x))) (dw)/(dy) (h(x)) (dy)/(dx) (x) = f^prime(g(h(x))) g^prime (h(x)) h^prime(x)#