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# Lesson 3-6 - PowerPoint PPT Presentation

Lesson 3-6. Implicit Differentiation. Objectives. Use implicit differentiation to solve for dy/dx in given equations Use inverse trig rules to find the derivatives of inverse trig functions. Vocabulary.

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### Lesson 3-6

Implicit Differentiation

• Use implicit differentiation to solve for dy/dx in given equations

• Use inverse trig rules to find the derivatives of inverse trig functions

• Implicit Differentiation – differentiating both sides of an equation with respect to one variable and then solving for the other variable “prime” (derivative with respect to the first variable)

• Orthogonal – curves are orthogonal if their tangent lines are perpendicular at each point of intersection

• Orthogonal trajectories – are families of curves that are orthogonal to every curve in the other family (lots of applications in physics (example: lines of force and lines of constant potential in electricity)

If a function (or a relation) can not be set into the for y = f(x), then implicit differentiation (differentiating both sides with respect to x and solving for y’) can be used to find the derivative.

Example: (a circle with radius 10, which is a relation and not a function) x² + y² = 100

dx dy dy

2x --- + 2y ---- = 0  2x + 2y ---- = 0

dx dx dx

dy dy -x

2x = -2y ----  ---- = -----

dx dx y

Example: (a circle with radius 10, which is a relation, not a function)

x² + y² = 100

If we tried solving for y first in the above circle we would get the following:

y = ± √100 - x² this gives us two functions y = √100 - x²and y = -√100 - x²

Differentiating the first (upper semi-circle) of these would give us:

dy -2x -x -x

---- = ½ (100 - x²)-½ (-2x) = ----------------- = -------------- = --------

dx 2√100 - x²√100 - x²y

solving for the second one (lower semi-circle) again gives us the same answer because y is negative for all values in it.

• Differentiate both sides of the equation with respect to x.

• Collect all terms involving dy/dx on one side of the equation and move all other terms to the other side.

• Factor dy/dx out of the terms on the one side.

• Solve for dy/dx by dividing both sides of the equation by the factored term.

Find the derivatives of the following:

1. y³ + 7y = x³

2. 4x²y – 3y = x³ – 1

3y² (dy/dx) + 7(dy/dx) = 3x²

dy/dx(3y² + 7) = 3x²

dy/dx = 3x² / (3y² + 7)

8xy + 4x² (dy/dx) - 3(dy/dx) = 3x²

dy/dx(4x² - 3) = 3x² - 8xy

dy/dx = (3x² - 8xy) / (4x² - 3)

Find the derivatives of the following:

3. x² + 5y³ = x + 9

4. Find Dty if t³ + t²y – 10y4 = 0

2x + 15y²(dy/dx) = 1

dy/dx(15y²)= 1 – 2x

dy/dx = ( 1 – 2x) / (15y²)

3t² + 2ty + t²(dy/dx) – 40y³(dy/dx) = 0

dy/dx(t² - 40y³) = 3t² + 2ty

dy/dx = (3t² + 2ty) / (t² - 40y³)

Find the derivatives of the following:

y – 1 = 1/3(x – 0)

5. Find the equation of the tangent line to the curve y³ – xy² + cos(xy) = 2 at x = 0.

6. Find dy/dx at (2,1) if 2x²y – 4y³ = 4.

y³ - (0)y² + cos(0) = 2  y³ = 1 y = 1

3y²(dy/dx) – y² - 2xy(dy/dx) – sin(xy) (xdy/dx + y) = 0

dy/dx(3y² - 2xy – xsin(xy)) = y² + ysin(xy)

dy/dx = (y² + ysin(xy)) / (3y² - 2xy – xsin(xy)) = (1)/ (3) = 1/3

2x²(dy/dx) + 4xy – 12y² (dy/dx) = 0

dy/dx(2x² + 12y²) = -4xy

dy/dx = (-4xy) / (2x² + 12y²) = (-4(2)(1)) / (22² + 12(1)) = -8/20

Find the equation of the normal line (line perpendicular to the tangent line) to the curve 8(x² + y²)² = 100(x² – y²) at the point (3,1).

8(x² + y²)² = 100(x² - y²)

16(x² + y²) (2x + 2y(dy/dx)) = 200x – 200y (dy/dx)

32x³ + 32xy² + 32x²y(dy/dx) + 32y³(dy/dx) = 200x – 200y(dy/dx)

(dy/dx)(32x²y + 32y³ + 200y) = 200x – 32x³ - 32xy²

dy/dx = (200x – 32x³ - 32xy²) / (32x²y + 32y³ + 200y)

dy/dx = (600 – 32(27) – 32(3)) / (32(9) + 32 + 200) = -9/13

Normal line slope, -1/mt = 13/9 y – 1 = (13/9)(x – 3)

• Summary:

• Use implicit differentiation when equation can’t be solved for y = f(x)

• Derivatives of inverse trig functions do not involve trig functions

• Homework:

• pg 233-235: 1, 6, 7, 11, 17, 25, 41, 47