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Math 103, week 6 notes

by: Cambria Revsine

Math 103, week 6 notes MATH 103 001

Cambria Revsine

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About this Document

These notes cover material from chapters 3.6-3.8, from Thomas' Calculus
Intermediate Algebra Part III
William Simmons
Class Notes
Math, Calculus
25 ?




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This 5 page Class Notes was uploaded by Cambria Revsine on Sunday February 28, 2016. The Class Notes belongs to MATH 103 001 at University of Pennsylvania taught by William Simmons in Spring 2016. Since its upload, it has received 29 views. For similar materials see Intermediate Algebra Part III in Mathematics (M) at University of Pennsylvania.

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Date Created: 02/28/16
Math 103—Week 6 Notes—3.6­3.8 3.6: The Chain Rule: When differentiating a composite function, find the derivative of the outside function first,  keeping the original inside function(s), and then multiply by the derivative of the inside  function(s)   Composite functions: f   g = f(g(x))  y= f (u) u=g(x) Derivative:  f '(g(x))∙g'(x) In other terms… dy dy du = ∙ dx du dx Ex: Find the derivative of  y=(3x +1) 2 2 y= f (u)=u u=g (x=3x +1 f '(u)∙g'(x) 2u∙6x ¿2 3x +1 )∙6x 3 ¿36x +12x Ex: 2 x Find the derivative of  sin(x +e )  with respect to x (¿x +e )=cos? (x +e )∙(2x+e )x d sin¿ dx 3.7: Implicit Differentiation: Implicit Differentiation is used when differentiating a function with x and y, where it is difficult or impossible to manipulate it into a y=x  form  2 2 2 Examples:    y −x=0       x +y =25   dy 1. Differentiate both sides of the equation with respect to x. The derivative of y dx   or y' . 2. Collect the terms with  dy  on one side of the equation and solve for dy . dx dx dy 2 2 Ex: Find  dx  of  x +y =25 x (¿¿2) dy 2 dy dy + dx (y = dx (25 ) ¿ dx dy 2x+2y dx =0   dy 2y dx =−2x    dy =x dx y dy Ex: Find   of  y =x +sin ( xy¿ dx 2y y=2x+cos (xy)(xy +y) ' ' 2y y−cos (xy xy =2x+cos (xy)y xy )x 2y−cos¿=2x+cos (xy)y y'¿ ' 2x+ycos (xy) y = 2y−xcos xy( ) **To find the second derivative, follow normal derivative rules to differentiate the first  derivative ( y'' ¿ , and then plug in the first derivative into thy'  values of the second  derivative  Normal line: Line perpendicular to the tangent line of a point (take negative reciprocal of the  derivative to find the slope of the normal line)  3 3 Ex: Show that the point (2,4) lies on the curvex +y −9xy=0 . Then find the tangent and  normal lines to the curve. To show the point lies on the curve, plug in the point into x and y: 2 +4 −9 (2)(4)=0  Find derivative: 3x +3y y −9(xy +y)=0 2 ' ' 2 3y y −9x y =−3x +9y y'3y −9x =−3x +9y 2 y =−3x +9y 3y −9x 2 ' 3y−x y = 2 y −3x Plug in point into the derivative to find the tangent slope: ' 3(4)−(2) 2 y = 2 (4) −3(2) ' 4 y = 5 4 12 4= ( )+b;b= 5 5 4 12 Tangent line:  y= x+ 5 5 Find normal line slope: −5 m= 4 −5 13 4= ( )b;b= 4 2 −5 13 Normal line:  y= x+ 4 2 3.8: Derivatives of Inverse Functions and Logarithms: f The Derivative Rule for Inverses:  (¿¿−1)'(x)= 1 f '( f1( )) ¿ −1 Ex:  f(x)=x −2 ,  x>0 . Find  d f  at  x=6= f (2). dx f (¿¿−1)'(6)= 1 f '( f ( ) ¿ ' 2 (f)=3x f−1 6 =2 f 2 ¿ ¿ ¿2 3¿ (¿¿−1)'(6)= 1 ¿ ¿ Derivative of the Natural Log Function: d lnu= 1 du dx x dx Ex: Find  d ln(x +3) dx 1 2x 2 •2x= 2 x +3 x +3 u Derivative of  a : d u u du dx a =a lna dx Derivative of  logau : d 1 du log a= dx ulna dx n nlnx ** x =e ln x =nlnx x Ex: Differentiate  f (x)=x ** x =e xlnx f (x)= d (exln) dx ¿exlnx xlnx) dx xlnx 1 ¿e (nx+x• x) x ¿x (lnx+1) e as a Limit: x 1+¿ ¿ ¿ e=lim ¿ x→ 0


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