How do you find the equation of the tangent line to the curve at the given point #sin(x+y)=2x-2y# at point (pi, pi)?

1 Answer
Aug 24, 2016

The equation is #y = 1/3x + (2pi)/3#.

Explanation:

Start by expanding #sin(x + y)# using the sum formula #sin(a + b) = sinacosb + sinbcosa#

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

#sinxcosy + sinycosx = 2x - 2y#

Use a combination of implicit differentiation and the product rule:

#cosxcosy - sinxsiny(dy/dx) + cosycosx(dy/dx) - sinysinx = 2 - 2(dy/dx)#

#cosycosx(dy/dx) - sinxsiny(dy/dx) + 2(dy/dx) = 2 + sinxsiny - cosxcosy#

#dy/dx(cosycosx - sinxsiny + 2) = 2 + sinxsiny - cosxcosy#

#dy/dx = (2 + sinxsiny - cosxcosy)/(cosxcosy - sinxsiny + 2)#

We can find the slope of the tangent by inserting our point, #(x, y)# into the derivative. Here, the point is #(pi, pi)#

#m_"tangent" = (2 + 0(0) - (-1 xx -1))/(-1 xx -1 - 0 xx 0 + 2)#

#m_"tangent" = 1/3#

We can now use point-slope form to determine the equation of the tangent.

#y - y_1 = m(x - x_1)#

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

#y - pi = 1/3x - pi/3#

#y = 1/3x - pi/3 + pi#

#y = 1/3x + (2pi)/3#

So, the equation of the tangent is #y = 1/3x + (2pi)/3#.

Hopefully this helps!