`x^2 + 2xy - y^2 + x = 2, (1, 2)` 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 28 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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hkj1385 | (Level 1) Assistant Educator

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

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

Now, the given function is :-

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

Differentiating both sides w.r.t 'x' we get;

2x + 2x(dy/dx) + 2y - 2y(dy/dx) + 1 = 0

Putting x =1 & y =2 in the above equation we get

slope of the tangent at (1,2) = dy/dx  at (1,2) = 7/2

Now, equation of the tangent to the given curve at (1,2) is :-

y - 2 = (7/2)*(x-1)

or, 2y - 4 = 7x - 7

or, 2y = 7x - 3 = equation of the tangent at point (1,2)

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