Find as directed the equation of the tangent line to the cur

Find, as directed, the equation of the tangent line to the curve implicitly defined as a function of x.y + 8ey = 8 + In(x) at the point(1,0) y2 - e2x = ex In(y) at the point (0,1)

Solution

a) differentiating,

dy/dx(1 + 8e^y) = 1/x

at (1,0)

dy/dx(1 +8) = 1/1

dy/dx = 1/9

hence, equation of line is :

y = 1/9(x-1)

x-9y = 1

b) differentiating,

2y(dy/dx) - 2e^(2x) = e^y((dy/dx)/y) + lny*e^x

(dy/dx)(2y- e^y/y) = 2e^(2x) + lny *e^x

at (0,1)

dy/dx(2-e) = 2 + 0

dy/dx = 2.784

hence, equation of line is :

y-1 = 2.783*(x-0)

2.783x - y + 1 = 0