Use implicit differentiation to find an equation of the tangent line to the curve sin(x+y)=8x−8y at the point (π,π)

Question

Please help I can't seem to figure this out

Answer 1

cos(x+y) (1 + y') = 8 - 8y'

cos(x+y)+ cos(x+y)y' = 8 - 8y'

cos(x+y) - 8 = - 8y' - cos(x+y)y'

cos(x+y) - 8 = - y' (8 + cos(x+y))

the derivative is:

8 - cos(x+y)

-------------

8 + cos(x+y)

at (pi,pi), the slope is

M=

8 - cos(2*pi)

----------------

8 + cos(2*pi)

=

8 - 1

-------

8 + 1

= 7/9

B = y - mx

= pi - (7/9) pi

= (2/9) pi

y = (7/9)x + (2/9)pi

(1/9) (7x + 2*pi)

Answer 2

Differentiate through wrt x

cos (x+y) d/dx (x+y) = 8x - 8 dy/dx

cos (x+y) (1 + dy/dx) = 8x - 8 dy/dx

x=π, y=π

1 + dy/dx = 8π - 8 dy/dx

dy/dx = (8π - 1) / 9

This is the slope of the tangent line

y = mx + b

m = (8π - 1) / 9

π = π (8π-1)/9 + b

b = π - π (8π-1)/9

y = (8π-1)x / 9 + π - π(8π-1)/9