16
Implicit Differentiation

Lesson 6 Nov 23 24 09

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Page 1: Lesson 6 Nov 23 24 09

Implicit Differentiation

Page 2: Lesson 6 Nov 23 24 09

y = 5x + 1Derivative?

Explicit differentiation

Having the dependent variable expressed directly in terms of the independent variable.

Page 3: Lesson 6 Nov 23 24 09

Given y2 ­ 2x = y

Find y '

Step 1. Take derivative of each sideddx( )y2 ­ 2x      = d

dx( )y

dydx= 22y ­ 1

Page 4: Lesson 6 Nov 23 24 09

Step 2. Put all terms with       on the same sidedydx

Step 3. Factor out  dydx

Page 5: Lesson 6 Nov 23 24 09

Step 4. Solve for  dydx

Page 6: Lesson 6 Nov 23 24 09

y2 ­ 2x = y

Find the slope of the tangent line at (1, ­1)

Page 7: Lesson 6 Nov 23 24 09

if x2 + y2 = 36Find  dydx

­xy

Page 8: Lesson 6 Nov 23 24 09

Find  dydx if   2x3 ­ 3y3 = 4xy

Page 9: Lesson 6 Nov 23 24 09

Product Rule

ddx[ f(x) g(x). ] = f '(x).g(x)f(x) . g '(x) +

Page 10: Lesson 6 Nov 23 24 09

Find  dydx x y8 + 6x  =  3y

4y + xy6xy3 ­ 4x

Page 11: Lesson 6 Nov 23 24 09
Page 12: Lesson 6 Nov 23 24 09

Find  dydx

cos y + x3 = sin x

3x2 ­ cos xsin y

Page 13: Lesson 6 Nov 23 24 09

3e5x + 6ey2 = 8Find  dy

dx­5e5x

4yey2

Page 14: Lesson 6 Nov 23 24 09
Page 15: Lesson 6 Nov 23 24 09

What is the slope of the tangent line to 

5x2 +3xy + 2y2 + 12y ­ 5 = 0

at the point (­1, 0)

Page 16: Lesson 6 Nov 23 24 09

exercise 4.5

Questions 1 ­ 23