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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

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**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**• 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)**Implicit Differentiation**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**Implicit Differentiation**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.**Guidelines for Implicit Differentiation**• 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.**Example 1**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)**Example 2**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)**Example 3**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**Example 4**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 & Homework**• 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

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