Question

(A) Sin(y/x) +ye^(2-x) = x-y (B) √y - 4ln x = cos (x+y) (C) e^(2y) + tan(1/x) =(x²/y) (D) ln y+2x =(3-y)^3 (E)xy²-sin(x+2y)=2x . Find the dy/dx (implicit differentiation)?

Answer 1

To keep the answer at a reasonable length, I will only show you parts (A) and (E).

Explanation:

So, for part (A) you have

`sin(y/x) + ye^(2-x) = x- y`

To get `(dy)/dx`, you need to use implicit differentiation. Differentiate both sides with respect to `x`

`d/dx[sin(y/x) + ye^(2-x)] = d/dx(x-y)`

To make the calculations easier to follow, I'll solve each of these derivatives separately. First, you have

`d/dx[sin(y/x)] = cos(y/x) * d/dx(y * x^(-1))`

`d/dx[sin(y/x)] = cos(y/x) * [(dy)/dx * x^(-1) + y * (-x)^(-2)]`

Next

`d/dx(ye^(2-x)) = (dy)/dx * e^(2-x) + y * d/dx(e^(2-x))`

`d/dx(ye^(2-x)) = (dy)/dx * e^(2-x) + y * e^(2-x) * (-1)`

`d/dx(ye^(2-x)) = (dy)/dx * e^(2-x) - y * e^(2-x)`

Plug these back into your target calculation to get

`cos(y/x) * (dy)/dx * x^(-1) + cos(y/x) * (-yx^(-2)) + (dy)/dx * e^(2-x) - y * e^(2-x) = 1 - (dy)/dx`

Isolate `(dy)/dx` on one side of the equation to get

`cos(y/x) * (dy)/dx * x^(-1) + (dy)/dx * e^(2-x) + (dy) = 1 + y * e^(2-x) +yx^(-2) * cos(y/x)`

`(dy)/dx * [x^(-1)cos(y/x) + e^(2-x) + 1] = 1 + y * e^(2-x) +yx^(-2) * cos(y/x)`

Finally, you have

`(dy)/dx = [1 + y * e^(2-x) +yx^(-2) * cos(y/x)]/[1 + e^(2-x) - x^(-1)cos(y/x)]`

You can simplify this to the form

`(dy)/dx = (x^2 + y * x^2 * e^(2-x) + ycos(y/x))/x^color(red)(cancel(color(black)(2))) * color(red)(cancel(color(black)(x)))/(x + x * e^(2-x) - cos(y/x))`

`(dy)/dx = color(green)((x^2 + y x^2e^(2-x) + ycos(y/x))/(x[x + xe^(2-x) - cos(y/x)])`

For part (E) you have

`xy^2 - sin(x+ 2y) = 2x`

The exact same approach applies here as well. Differentiate both sides with respect to `x`

`d/dx[xy^2 - sin(x + 2y)] = d/dx(2x)`

This will get you

`d/dx(xy^2) = d/dx(x) * y^2 + x * (dy)/dx * 2y`

`d/dx(xy^2) = y^2 + 2xy(dy)/dx`

and

`d/dx[sin(x + 2y)] = cos(x + 2y) * d/dx(x + 2y)`

`d/dx[sin(x + 2y)] = cos(x + 2y) * [d/dx(x) + 2(dy)/dx]`

`d/dx[sin(x + 2y)] = cos(x + 2y) * (1 + 2(dy)/dx)`

Plug these back into your target calculation to get

`y^2 + 2xy(dy)/dx - [cos(x + 2y) + 2 cos(x + 2y)(dy)/dx] = 2`

`y^2 + 2xy(dy)/dx - cos(x + 2y) - 2cos(x + 2y)(dy)/dx = 2`

Once again, isolate `(dy)/dx` on one side

`(dy)/dx * 2[xy - cos(x + 2y)] = 2 - y^2 + cos(x + 2y)`

`(dy)/dx = color(green)((2 - y^2 + cos(x + 2y))/(2[xy - cos(x + 2y)])`