For more information, please see full course syllabus of College Calculus: Level I

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For more information, please see full course syllabus of College Calculus: Level I

For more information, please see full course syllabus of College Calculus: Level I

### 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 = x

^{3}- [d/dx]2y = [d/dx]x
^{3} - 2 [dy/dx] = 3 x
^{2}

[dy/dx] = [3/2] x

^{2}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 e

^{y}= x- [d/dx]e
^{y}= [d/dx]x - e
^{y}[dy/dx] = 1

[dy/dx] = e

^{−y}Find [dy/dx] given lny = x

^{2}+ cosx- [d/dx]lny = [d/dx]x
^{2}+ [d/dx]cosx - [1/y] [dy/dx] = 2x − sinx

[dy/dx] = y(2x − sinx)

Find [dy/dx] given √{y

^{2}+ y} + [1/x] = 1- [d/dx]√{y
^{2}+ y} + [d/dx][1/x] = [d/dx]1 - [1/2] [1/(√{y
^{2}+ y})] [d/dx](y^{2}+ y) + (− [1/(x^{2})]) = 0 - [1/(2 √{y
^{2}+ y})] (2y [dy/dx]+ [dy/dx]) − [1/(x^{2})] = 0 - [1/(2 √{y
^{2}+ y})] (2y [dy/dx]+ [dy/dx]) = [1/(x^{2})] - [dy/dx]([(2y + 1)/(2 √{y
^{2}+ y})]) = [1/(x^{2})]

[dy/dx] = [(2 √{y

^{2}+ y})/(x^{2}(2y + 1))]Find [dy/dx] given [(x

^{2})/4] + [(y^{2})/9] = 4- [d/dx][(x
^{2})/4] + [d/dx][(y^{2})/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 + e

^{y}= cosx- [d/dx]lny + [d/dx]e
^{y}= [d/dx]cosx - [1/y] [dy/dx]+ e
^{y}[dy/dx] = − sinx - [dy/dx]([1/y] + e
^{y}) = − sinx

[dy/dx] = − [sinx/([1/y] + e

^{y})]Find [dy/dx] given y

^{2}+ 3y = cos^{−1}x^{3}- [d/dx]y
^{2}+ [d/dx]3y = [d/dx]cos^{−1}x^{3} - 2y [dy/dx]+ 3 [dy/dx] = − [1/(1 − (x
^{3})^{2})] [d/dx]x^{3} - [dy/dx](2y + 3) = − [(3x
^{2})/(1 − x^{6})]

[dy/dx] = [(−3x

^{2})/((2y + 3)(1 − x^{6}))]Find [dy/dx] given x

^{2}+ (y + 3)^{2}= 1- [d/dx]x
^{2}+ [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} = x

^{2}- [d/dx]√{2y + siny} = [d/dx]x
^{2} - [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
- 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
- Additional Example 5
- Additional Example 6

0 answers

Post by Zhe Yang on November 3, 2014

you did a wonderful job explaining it, thanks!

0 answers

Post by kathy park on December 16, 2012

Would 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 answers

Post by David Jarrett on June 26, 2012

Wow, thanks so much! This totally cleared everything up for me!

0 answers

Post by Ron Weldy on March 2, 2011

The 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!