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• ## Related Books

 0 answersPost by Joyce Ferreira on October 27 at 07:31:52 PMDear Prof. Switkes,It has been a while that I took a calculus class and I need to review partial differentiation, so I can be able to follow the lectures on my physical chemistry class. Please, I would liek to ask you what calculus lecture should I watch? Thank you. Best regards. 0 answersPost by Zhe Yang on November 3, 2014you did a wonderful job explaining it, thanks! 0 answersPost by kathy park on December 16, 2012Would you please check the second problem ? If you use product rule,d/dx[f(x)g(x)] = f'(x)gx) + f(x)g'(x), the solution is d/dx[4x^3*y^8] = 12x^2*y^8 + 4x^3*8y^7dy/dx = 4x^2y^7(3y + 8xdy/dx) 0 answersPost by David Jarrett on June 26, 2012Wow, thanks so much! This totally cleared everything up for me! 0 answersPost by Ron Weldy on March 2, 2011The problems we have at school are like cos(y)=x; (0,pie/2) or sin(y)=5x^4-5; (1,pie)I don't know I am stuck!

### Implicit Differentiation

• This method allows you to determine the derivative as a function of both and in situations in which it is not convenient to solve explicitly for as a function of .
• The key to these problems is to recognize that and to use the Chain Rule whenever appears.
• You will be differentiating both sides of an equation with respect to .
• You will often find yourself using the Chain Rule within a Product Rule computation.

### 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.

### 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
• Purpose 0:09
• Implicit Function
• Lecture Example 1 0:32
• Lecture Example 2 7:14
• Lecture Example 3 11:22
• Lecture Example 4 16:43