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# Implicit Differentiation - PowerPoint PPT Presentation

Implicit Differentiation. Lesson 3.5. Introduction. Consider an equation involving both x and y: This equation implicitly defines a function in x It could be defined ex plicitly. Differentiate. Differentiate both sides of the equation each term one at a time

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## PowerPoint Slideshow about 'Implicit Differentiation' - arnaud

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Presentation Transcript

### Implicit Differentiation

Lesson 3.5

• Consider an equation involving both x and y:

• This equation implicitly defines a function in x

• It could be defined explicitly

• Differentiate both sides of the equation

• each term

• one at a time

• use the chain rule for terms containing y

• For we get

• Now solve for dy/dx

• Then gives us

• We can replace the y in the results with the explicit value of y as needed

• This gives usthe slope on the curve for any legal value of x

• Given x3 + y3 = y + 21find the slope of the tangent at (3,-2)

• 3x2 +3y2y’ = y’

• Solve for y’

Substitute x = 3, y = -2

Second Derivative

• Given x2 –y2 = 49

• y’ =??

• y’’ =

Exponential & Log Functions

• Given y = bx where b > 0, a constant

• Given y = logbx

• Given

• Take the log of both sides, simplify

• Now differentiate both sides with respect to x, solve for dy/dx

• On older TI calculators, you can declare a function which will do implicit differentiation:

• Usage: