2. The Problem  Complex Functions  Why?  not all derivatives can be found through the use of the power, product, and quotient rules

3. 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

4. Working To A Solution cont.  Therefore,  (∆y/∆x) = (∆y/∆z) (∆z/∆x)  Since (dy/dx)=limx  0 (∆y/∆x)  (dy/dx) = (dy/dz) (dz/dx)  This is known as “The Chain Rule”

5. 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

6. 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

7. Example f(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)

8. Example f(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)

9. Homework  Chapter 3.4 Problems: 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.

10. Sources  Slideshow created using Microsoft PowerPoint  Clipart and themes supplied through Microsoft PowerPoint  Mathematics reference and notation from “Calculus: Single and Multivariable” 4 th edition, by Hughes- Hallet|Gleason|McCallum|et al.