Overview
Occasionally, you will have what are called composite functions; that is, functions that are composed of multiple functions and thus cannot be differentiated easily. In reality, although these may look tricky, they are actually fairly straightforward.
Function composed of two functions
Suppose we have an equation written as f(g(x)). Then
d/dx f(g(x)) = f’(g(x))g’(x)
If the equation is more complicated, such as
d/dx(f(g(h(x))),
then it can be broken thought of as doing the chain rule multiple times.
Think of it instead as f(j(x)) where j(x) = g(h(x)), and as we know the derivative of f(j(x)) is f'(j(x))j'(x). So since j(x)=g(h(x)), we can rewrite this as f'(g(h(x)))*(g(h(x)))’, or f'(g(h(x)))g'(h(x))h'(x).
It can be thought of as a chain, in that first we take the derivative of the outer equation and then move steadily inward.
Examples
d/dx(sin(2x))
Then f(y)=sin(y) and g(x)=2x, so sin(2x) can be written as f(g(x)).
Then as explained before, d/dx(f(g(x)) = f’(g(x))g’(x)
=(sin(g(x)))’(2x)’
=cos(g(x))*2
=2cos(2x)
d/dx(3(cos(2x3))4)
=3d/dx(cos(2x3))4)
Let f(z)=z4, g(y)=cos(y), and h(x)=2x3
Then f’(z)=4z3
g’(y)=-sin(y)
h’(x)=6x2
And since the equation can be written as f(g(h(x))), then
3*d/dx(f(g(h(x))))=3*f’(g(h(x)))*g’(h(x))*h’(x)
Or, by substituting the values in,
3*4(cos(2x3))3(-sin(2x3)*6x2
And, by combining like terms and
=-72x2sin(2x3)(cos(2x3))3
Interested in calculus tutoring services? Learn more about how we are assisting thousands of students each academic year.
SchoolTutoring Academy is the premier educational services company for K-12 and college students. We offer tutoring programs for students in K-12, AP classes, and college. To learn more about how we help parents and students in Cumberland, MD: visit Tutoring in Cumberland, MD