Implicit Differentiation Practice A | Answer Key- Edubirdie

University:
Southwest Minnesota State University
Course:
MATH 140 | Calculus: A Short Course
Academic year:

2023
Views:

497

Pages:

2
Author:

Richard Petersen

Implicit Differentiation PracticeA - answer 1. Find the derivative of the equation x^2 + y^2 = 25 with respect to x. Answer: d/dx (x^2 + y^2) = d/dx (25) 2x + 2y * dy/dx = 0 dy/dx = -x/y 2. Find the derivative of the equation y^2 = x^3 - x with respect to x. Answer: d/dx (y^2) = d/dx (x^3 - x) 2y * dy/dx = 3x^2 - 1 dy/dx = (3x^2 - 1)/2y 3. Find the equation of the tangent line to the curve x^2 + xy + y^2 = 7 at the point (1,2). Answer: Taking the derivative of the equation implicitly, we get: 2x + y + xy' + 2yy' = 0 Substituting (1,2), we get: 2(1) + 2 + 1y' + 2(2)y' = 0 5y' = -4 y' = -4/5 Using the point-slope form, the equation of the tangent line is y - 2 = (-4/5)(x - 1). 4. Find the derivative of the equation ln(xy) = x^2 + y^2 with respect to x. Answer: Taking the derivative of the equation implicitly, we get: (1/y) + (x/y) * dy/dx = 2x dy/dx = (2xy - y)/x 5. Find the equation of the tangent line to the curve y^3 + x^2 = 10y at the point (1,2). Answer: Taking the derivative of the equation implicitly, we get: 2x + 3y^2 * dy/dx = 10 Substituting (1,2), we get: 2 + 3(2)^2 * dy/dx = 10 dy/dx = (10-2)/12 = 2/3 Using the pointslope form, the equation of the tangent line is y - 2 = (2/3)(x 1).

Description

Related Documents

Struggling with your assignment and deadlines?

Free up your schedule!

Take 5 seconds to unlock