130 likes | 248 Views
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))
E N D
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.
Sources • Slideshow created using Microsoft PowerPoint • Clipart and themes supplied through Microsoft PowerPoint • Mathematics reference and notation from “Calculus: Single and Multivariable” 4th edition, by Hughes-Hallet|Gleason|McCallum|et al.