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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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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 explicitly
Differentiate • 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
Differentiate • 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 View Spreadsheet Example
Slope of a Tangent Line • Given x3 + y3 = y + 21find the slope of the tangent at (3,-2) • 3x2 +3y2y’ = y’ • Solve for y’ Substitute x = 3, y = -2
Substitute Second Derivative • Given x2 –y2 = 49 • y’ =?? • y’’ =
Note: this is a constant Exponential & Log Functions • Given y = bx where b > 0, a constant • Given y = logbx
Using Logarithmic Differentiation • Given • Take the log of both sides, simplify • Now differentiate both sides with respect to x, solve for dy/dx
Implicit Differentiation on the TI Calculator • On older TI calculators, you can declare a function which will do implicit differentiation: • Usage: Newer TI’salready havethis function
Assignment • Lesson 3.5 • Page 171 • Exercises 1 – 81 EOO