INSTRUCTORS Raffi Hovasapian John Zhu

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• ## Related Books

 3 answersLast reply by: Arshin JainSun Dec 15, 2013 11:22 PMPost by Steve Denton on October 11, 2012On 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 answerLast reply by: Steve DentonThu Oct 11, 2012 7:59 PMPost by James Xie on July 5, 2012For 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)-13)(dy/dx) = (2x(y^2)-1)/(cos(y)-2(x^2)y)Would this be correct too? (Sorry if it looks unclear here) 1 answerLast reply by: Joshua SpearsTue Feb 26, 2013 1:06 PMPost by James Xie on July 5, 2012For 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.