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Warm-Up: Find f’(x) if f(x)=(3x 2 -6x+2) 3

Warm-Up: Find f’(x) if f(x)=(3x 2 -6x+2) 3. Section 6.4: Implicit Differentiation. Objective: To take the derivative of implicitly defined functions. Almost all the functions we have worked with so far have been of the form y = f(x) In these cases, y is given explicitly in terms of x

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Warm-Up: Find f’(x) if f(x)=(3x 2 -6x+2) 3

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  1. Warm-Up: Find f’(x) if f(x)=(3x2-6x+2)3

  2. Section 6.4: Implicit Differentiation Objective: To take the derivative of implicitly defined functions.

  3. Almost all the functions we have worked with so far have been of the form y = f(x) In these cases, yis given explicitly in terms of x Examples: y = 2x + 5 ,y = x2 + x + 6 , f(x)=(3x2-6x+2)3

  4. 5xy – 4x = 2 Is an implicit function. However, it can easily be solved for y. 5xy = 2 + 4x 5xy = 2 + 4x 5x5x y = 2 + 4x 5x

  5. Not all implicit functions can be rewritten explicitly • Example: y5 + 7y3 + 6x2y2 + 4yx3 + 2 = 0 In such cases, it is possible to find the derivative, dy/dx by a process called implicit differentiation.

  6. Implicit Differentiation • We assume y is a function defined in terms of x • We differentiate using the chain rule: • Variables disagree • Derivative of inner function is implicitly defined

  7. Implicit Differentiation To find dy/dx for an equation containing x and y: • Differentiate on both sides of the equation with respect to x, keeping in mind that y is assumed to be a function of x. • When differentiating x terms, take derivative as usual • When differentiating y terms, you assume y is implicitly defined as a function of x. Use chain rule. 2. Place all terms with dy/dx on one side of the equal sign, and all terms without dy/dx on the other side. 3. Factor out dy/dx, and then solve for dy/dx.

  8. Examples Find the derivative of the following functions. 1. y3 + y2 – 5y – x2 = -4 • x2 – 2xy + y3 = 5 • sinx+ x2y = 10 • 6x2 + 3xy+ 2y2 + 17y – 6 = 0

  9. More Fun 6. Find the slope of the curve y2 = 5x4 – x2 at the point (1, 2) 7. Find the line that is tangent to the curve at the given point. y2 – 2x – 4y – 1 = 0; (-2, 1)

  10. Find the value of x3y= 2x+3y2 at the point (0,2).

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