Explicit/Implicit Explicit functions: y = 3x – 2 Implicit functions:

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Presentation transcript:

Explicit/Implicit Explicit functions: y = 3x – 2 Implicit functions: y2 + 2yx 4x2 = 0 y5 - 3y2x2 + 2 = 0

Why Implicit Differentiation? When an applied problem involves an equation not in explicit form, implicit differentiation is used to locate extrema or to find rates of change.

Process for Implicit Differentiation To find dy/dx Differentiate both sides with respect to x (y is assumed to be a function of x, so d/dx) Collect like terms (all dy/dx on the same side, everything else on the other side) Factor out the dy/dx and solve for dy/dx

Example Find dy/dx if 3xy + 4y2 = 10 Differentiate both sides with respect to x: Use the product rule for (3x)(y) (The derivative of y is dy/dx)

Example Con’t Since y is assumed to be some function of x, use the chain rule for 4y2

Example Con’t Combine like terms

Example Con’t Factor and solve

Tangent Lines Find the equation of the tangent line to the curve below at the point (2,4)

Tangent Example x3 + y3 = 9xy Now what?

Tangent Line Con’t The slope of the tangent line is found by plugging (2,4) into the derivative Since we’re looking for the tangent line:

6.5 – Related Rates Time is often present implicitly in a model. This means that the derivative with respect to time must be found implicitly.

Related Rates Suppose x and y are both functions of t (time), and that x and y are related by xy2 + y = x2 + 17 When x = 2, y = 3, and dx/dt = 13. Find the value of dy/dt at that moment

Continued xy2 + y = x2 + 17 Product and chain rules Now substitute x = 2, y = 3 and dx/dt = 13

Continued Solve for dy/dt

Solving the Problems

Area Example A small rock is dropped into a lake. Circular ripples spread out over the surface of the water, with the radius of each circle increasing at a rate of 3/2 ft. per second. Find the rate of change of the area inside the circle formed by a ripple at the instant r = 4ft.

Area Example Area and radius are related by Take the derivative of each side with respect to time Since the radius is increasing at the rate of 3/2 ft per second

Area Continued Find the rate of change of the area inside the circle formed by a ripple at the instant r = 4ft.

One More--Volume A cone-shaped icicle is dripping from the roof. The radius of the icicle is decreasing at a rate of 0.2cm/hr, while the length is increasing at a rate of 0.8cm/hr. If the icicle is currently 4cm in radius, and 20cm long, is the volume of the icicle increasing or decreasing, and at what rate?

Volume The volume of a cone is found by A constant is being multiplied to 2 variables—use the product rule & chain rule

Volume The radius of the icicle is decreasing at a rate of 0.2cm/hr, while the length is increasing at a rate of 0.8cm/hr.

Volume The icicle is currently 4cm in radius and 20cm long. Plugging in we have: The volume is decreasing at a rate of about 20cm3/hr

6.6 Differentials: Linear Approximation

6.6 Differentials: Linear Approximation Find dy for y = 6x2 dy/dx = 12x dy = 12x dx

One More Find dy:

Linear Approximation Approximate

12.7 L’Hospital’s Rule We started this semester with a discussion of limits: On another note:

12.7 L’Hospital’s Rule Looking at This leads to a meaningless result and is called the indeterminate form

L’Hospital’s Rule L’Hospital’s rule gives us a quick way to decide whether a quotient with the indeterminate form has a limit. Do NOT use the quotient rule!

Example A quick check gives you 0/0 By L’H:

One More Find Substitution gives 0/0, so apply L’H Do it again!

Lab 3 Do these questions to hand in next time: 1. #54, P375 10. #15, P396 2. #1, P382 11. #9, P395 3. #4 & 7, P372 12. #9, P383 4. #3, P401 13. #23, P384 5. #28, P401 14. #19, P401 6. #20, P372 15. #29, P411 7. #32, P385 16. #43, P374 8. #3, P409 17. #26, P411 9. #11, P409 18-20. #36, 40, 45 P772