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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)=limx0 (∆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 ,[object Object],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, themes, and sound supplied by Microsoft PowerPoint Mathematics reference and notation from Hughes-Hallett, , Gleason, McCallum, and et al. Calculus: Single and Multivariable. 4th. 2005. 127-130. Print.

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The Chain Rule Powerpoint Lesson

  • 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.
  • 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. 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)
  • 8. 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)
  • 9. 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.
  • 10. Sources Slideshow created using Microsoft PowerPoint Clipart, themes, and sound supplied by Microsoft PowerPoint Mathematics reference and notation from Hughes-Hallett, , Gleason, McCallum, and et al. Calculus: Single and Multivariable. 4th. 2005. 127-130. Print.