THE CHAIN RULE

The Problem

Complex Functions

Why?

not all derivatives can be found through the use of the power, product, and quotient rules

Working To A Solution

Composite function f(g(x)) f is the outside function, g is the inside function

z=g(x), y=f(z), and y=f(g(x))

Therefore,

a small change in x leads to a small change in z

a small change in z leads to a small change in y

Working To A Solution cont.

Therefore,

(∆y/∆x) = (∆y/∆z) (∆z/∆x)

Since (dy/dx)=limx0 (∆y/∆x)

(dy/dx) = (dy/dz) (dz/dx)

This is known as “The Chain Rule”

Pushing Further

Looking back, z=g(x), y=f(z), and y=f(g(x))

Since (dy/dz)=f’(z) and (dz/dx)=g’(x) (d/dx) f(g(x)) = f’(z) × g’(x)

This allows us to rewrite the chain rule as(d/dx) f(g(x)) = f’(g(x)) × g’(x)

Therefore, the derivative of a composite function equals the derivative of the outside function times the derivative of the inside function

Example:f(x) = [ (x^3) + 2x + 1 ]^3

inside function = z = g(x) = (x^3) + 2x + 1 outside function = f(z) = z^3

g’(x) = (3x^2) + 2 f’(z) = 3[z]^2

(d/dx) f(g(x)) = 3 [ (3x^2) + 2 ] [ z]^2

= [ (9x^2) + 6 ] [ (x^3) + 2x + 1 ]^2

Examplef(x) = e^(4x^2)

inside function = z = g(x) = 4x^2 outside function = f(z) = e^z

g’(x) = 8x f’(z) = e^z

(d/dx) f(g(x)) = 8xe^z = 8xe^(4x^2)

Examplef(x) = [(x^3) + 1]^(1/2)

Inside function = z = g(x) = (x^3) + 1 outside function = f(z) = z^(1/2)

g’(x) = 3x^2 f’(z) = (1/2)z^(-1/2)

(d/dx) f(g(x)) = (3x^2) (1/2)z^(-1/2) = [(3/2)x^2] [(x^3) + 1]^(-1/2)

Homework

Chapter 3.4Problems: 1,2,4,5,6,7,8,10,11,12,15,16,18,20,27

Remember to show all work. Turn in the assignment before the beginning of the next class period.

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Microsoft PowerPoint Mathematics reference and notation from

“Calculus: Single and Multivariable” 4th edition, by Hughes-Hallet|Gleason|McCallum|et al.