Given the equation below, find dy/dx. 42x^4 + 9x^42y + y^5 = 52 dy/dx = (-168x^3 - 378yx^41) / (9x^42 + 5y^4) Now, find the equation of the tangent line to the curve at (1, 1). Write your answer in mx + b format y = -33x + 34 Question Help: Video
Added by John M.
Close
Step 1
Step 1:** Find the derivative of the curve using implicit differentiation: \[42x^4 + 9x^{42}y + y^5 = 52\] \[168x^3 + 378x^{41}y + 9x^{42}\frac{dy}{dx} + 5y^4\frac{dy}{dx} = 0\] \[9x^{42} + 5y^4\frac{dy}{dx} = -168x^3 - 378x^{41}y\] \[5y^4\frac{dy}{dx} = -168x^3 Show more…
Show all steps
Your feedback will help us improve your experience
Ma. Theresa Alin and 86 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
All parts of this problem refer to the function below. y = (4 + 2x)^(4/x) a) Use logarithmic differentiation to find dy/dx dy/dx = b) Find the slope of the tangent line at x = 1. Slope = c) Find the equation of the tangent line at x = 1. Tangent line: y =
Adi S.
Consider the curve defined by xy^2 - 2x^3 = 2 for y ≥ 0. a) Show that dy/dx = (6x^2 - y^2) / (2xy) b) Write an equation for the line tangent to the curve at the point (1,2). c) Find the x-coordinate of the point P at which the line tangent to the curve at P is horizontal. d) Find the value of d^2y/dx^2 at the point (1,2). If 3x^2 + 5x^2y^2 = 2y, then dy/dx =
Madhur L.
Find the equation of the tangent line at the given point on each curve. $$y+\frac{\sqrt{x}}{y}=3 ; \quad(4,2)$$
Applications of the Derivative
Implicit Differentiation
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD