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INSTRUCTORS Raffi Hovasapian John Zhu
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For more information, please see full course syllabus of Calculus AB
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Lecture Comments (8)

3 answers

Last reply by: Arshin Jain
Sun Dec 15, 2013 11:22 PM

Post by Steve Denton on October 11, 2012

On ex.5 at 16:00, adding the numerator terms requires multiplying the first term by sin y/sin y and therefore should be sin x sin^2 y. Right?

1 answer

Last reply by: Steve Denton
Thu Oct 11, 2012 7:59 PM

Post by James Xie on July 5, 2012

For the problem: x + sin(x) = (x^2)(y^2), this is the work I did:
1) 1 + cos(y)(dy/dx) = 2(x^2)y(dy/dx) + 2x(y^2)
2)(dy/dx)(cos(y)-2(x^2)y) = 2x(y^2)-1
3)(dy/dx) = (2x(y^2)-1)/(cos(y)-2(x^2)y)
Would this be correct too? (Sorry if it looks unclear here)

1 answer

Last reply by: Joshua Spears
Tue Feb 26, 2013 1:06 PM

Post by James Xie on July 5, 2012

For the second order example, why is it 4(8y+1)^2 in the beginning of the final answer? Shouldn't it start with 4(8y+1)?

Implicit Differentiation

  • Use when “y” cannot be extracted and isolated
  • Implicit differentiation method:
    • Take derivative with respective variables
    • Isolate to solve, usually by factoring than dividing all other terms
  • Second order implicit differentiation
    • Solve first order derivative
    • Solve second order derivative
    • Substitute term with first order derivative

Implicit Differentiation

Find [dy/dx] given 2y = x3
  • [d/dx] 2y = [d/dx] x3
  • 2 [dy/dx] = 3 x2
[dy/dx] = [3/2] x2
Find [dy/dx] given siny = x
  • [d/dx] siny = [d/dx] x
  • cosy [dy/dx] = 1
  • [dy/dx] = [1/cosy]
[dy/dx] = secy
Find [dy/dx] given ey = x
  • [d/dx] ey = [d/dx] x
  • ey [dy/dx] = 1
[dy/dx] = e−y
Find [dy/dx] given lny = x2 + cosx
  • [d/dx] lny = [d/dx] x2 + [d/dx] cosx
  • [1/y] [dy/dx] = 2x − sinx
[dy/dx] = y(2x − sinx)
Find [dy/dx] given √{y2 + y} + [1/x] = 1
  • [d/dx] √{y2 + y} + [d/dx] [1/x] = [d/dx] 1
  • [1/2] [1/(√{y2 + y})] [d/dx] (y2 + y) + (− [1/(x2)]) = 0
  • [1/(2 √{y2 + y})] (2y [dy/dx] + [dy/dx]) − [1/(x2)] = 0
  • [1/(2 √{y2 + y})] (2y [dy/dx] + [dy/dx]) = [1/(x2)]
  • [dy/dx] ([(2y + 1)/(2 √{y2 + y})]) = [1/(x2)]
[dy/dx] = [(2 √{y2 + y})/(x2(2y + 1))]
Find [dy/dx] given [(x2)/4] + [(y2)/9] = 4
  • [d/dx] [(x2)/4] + [d/dx] [(y2)/9] = [d/dx] 4
  • [2x/4] + [2y/9] [dy/dx] = 0
  • [2y/9] [dy/dx] = − [x/2]
[dy/dx] = − [9x/4y]
Find [dy/dx] given lny + ey = cosx
  • [d/dx] lny + [d/dx] ey = [d/dx] cosx
  • [1/y] [dy/dx] + ey [dy/dx] = − sinx
  • [dy/dx] ([1/y] + ey) = − sinx
[dy/dx] = − [sinx/([1/y] + ey)]
Find [dy/dx] given y2 + 3y = cos−1 x3
  • [d/dx] y2 + [d/dx] 3y = [d/dx] cos−1 x3
  • 2y [dy/dx] + 3 [dy/dx] = − [1/(1 − (x3)2)] [d/dx] x3
  • [dy/dx] (2y + 3) = − [(3x2)/(1 − x6)]
[dy/dx] = [(−3x2)/((2y + 3)(1 − x6))]
Find [dy/dx] given x2 + (y + 3)2 = 1
  • [d/dx] x2 + [d/dx] (y + 3)2 = [d/dx] 1
  • 2x + 2(y + 3) [d/dx] (y + 3) = 0
  • 2x + 2(y + 3) [dy/dx] = 0
  • 2(y + 3) [dy/dx] = −2x
  • [dy/dx] = [(−2x)/(2(y + 3))]
[dy/dx] = [(−x)/(y + 3)]
Find [dy/dx] given √{2y + siny} = x2
  • [d/dx] √{2y + siny} = [d/dx] x2
  • [1/2] [1/(√{2y + siny})] [d/dx] (2y + siny) = 2x
  • [1/2] [1/(√{2y + siny})] (2 [dy/dx] + cosy [dy/dx]) = 2x
  • [dy/dx] [(2 + cosy)/(2√{2y + siny})] = 2x
[dy/dx] = [(4x √{2y + siny})/(2 + cosy)]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Implicit Differentiation

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Implicit Differentiation: First Order 0:07
    • Example 1: Setting Up
    • Example 1: Solving
  • Implicit Differentiation: Second Order (Ex. 2) 4:55
  • Example 3: Implicit Differentiation 9:11
  • Example 4: Implicit Differentiation 9:56
  • Example 5: Implicit Differentiation With Double Derivative 12:46