# `sin(x + y) = 2x - 2y, (pi, pi)` Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

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hkj1385 | Certified Educator

*Note:- 1) If y = sinx ; then dy/dx = cosx ; *

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

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

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

*4) cos(2pi) = 1*

Now, the given function is :-

sin(x+y) = 2x - 2y

or, cos(x+y)*[1 + (dy/dx)] = 2 - 2*(dy/dx)

Thus, putting x = y = pi ; we get

cos(2pi)*[1 + (dy/dx)] = 2 - 2*(dy/dx)

or, [1 + (dy/dx)] = 2 - 2*(dy/dx)

or, dy/dx = 1/3 = slope of the tangent to the curve at (pi,pi)

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

y - pi = (1/3)*(x - pi)

or, 3y - 3pi = x - pi

or, 3y = x + 2pi = equation of the tangent