`x^2 + xy + y^2 = 3, (1, 1)` Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

Textbook Question

Chapter 3, 3.5 - Problem 27 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

1 Answer | Add Yours

hkj1385's profile pic

hkj1385 | (Level 1) Assistant Educator

Posted on

Note:- 1) If  y = x^n ; then dy/dx = n*x^(n-1) ; where n = real number

2) If y = u*v ; where both u & v are functions of 'x' , then

dy/dx = v*(du/dx) + u*(dv/dx)

3) If y = k ; where 'k' = constant ; then dy/dx = 0

Now, the given function is :-

(x^2) + xy + (y^2) = 0

or, 2x + y + x*(dy/dx) + 2y*(dy/dx) = 0

Thus, putting x = y = 1 ; we get

3 + 3*(dy/dx) = 0

or, 

or, dy/dx = -1 = slope of the tangent to the curve at (1,1)

Thus, equation of the tangent line to the given curve at the point (1,1) is :-

y - 1 = (-1)*(x - 1)

or, y -1 = -x + 1

or, x + y = 2 equation of the tangent

We’ve answered 318,928 questions. We can answer yours, too.

Ask a question