Question

Find the equation of the tangent line to the curve
2exy=x+y at (0,2).

Answer 1

The equation of the tangent line to the curve

3 x - y = 2

Explanation:

Step(i):-

Given function = f(x,y) =
`2e^(xy) -x-y=0` ...(i)

Differentiating equation (i) with respective to 'x' , we get

`2 e^(xy) (d)/(d x) (x y) -1 -(dy)/(dx) =0`

apply formula

`(d)/(dx) (UV) = UV^(l) +V U^(l)`

step(ii):-

⇒
`2 e^(xy) (x((d)/(d x) ( y))+y(1)) -1 -(dy)/(dx) =0`

⇒
`2 e^(xy) (x((d)/(d x) ( y))+ 2e^(xy) y(1)) -1 -(dy)/(dx) =0`

Taking common d y/d x

`(2 e^(xy) (x) -1)(dy)/(dx) =1- 2e^(x y) y(1))`

`(dy)/(dx) =(1- 2e^(x y) y(1)))/((2 e^(xy) (x) -1))`

put At (0,2)

`(dy)/(dx) =(1- 2e^(0) 2(1)))/((2 e^(0) (0) -1))=(1-4)/(-1) =3`

slope of the curve m = 3

Step(iii):-

The equation of the tangent line to the curve

`y-y_(1) =m(x-x_(1) )`

y - 2 = 3 ( x - 0 )

3 x - y = 2

Final answer:-

The equation of the tangent line to the curve

3 x - y = 2