(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)?

1 Answer
Aug 30, 2015

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- ysin(yx)+ye2x=xy

To get (dy)/dxdydx, you need to use implicit differentiation. Differentiate both sides with respect to xx

d/dx[sin(y/x) + ye^(2-x)] = d/dx(x-y)ddx[sin(yx)+ye2x]=ddx(xy)

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))ddx[sin(yx)]=cos(yx)ddx(yx1)

d/dx[sin(y/x)] = cos(y/x) * [(dy)/dx * x^(-1) + y * (-x)^(-2)]ddx[sin(yx)]=cos(yx)[dydxx1+y(x)2]

Next

d/dx(ye^(2-x)) = (dy)/dx * e^(2-x) + y * d/dx(e^(2-x))ddx(ye2x)=dydxe2x+yddx(e2x)

d/dx(ye^(2-x)) = (dy)/dx * e^(2-x) + y * e^(2-x) * (-1)ddx(ye2x)=dydxe2x+ye2x(1)

d/dx(ye^(2-x)) = (dy)/dx * e^(2-x) - y * e^(2-x)ddx(ye2x)=dydxe2xye2x

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)/dxcos(yx)dydxx1+cos(yx)(yx2)+dydxe2xye2x=1dydx

Isolate (dy)/dxdydx 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)cos(yx)dydxx1+dydxe2x+(dy)=1+ye2x+yx2cos(yx)

(dy)/dx * [x^(-1)cos(y/x) + e^(2-x) + 1] = 1 + y * e^(2-x) +yx^(-2) * cos(y/x)dydx[x1cos(yx)+e2x+1]=1+ye2x+yx2cos(yx)

Finally, you have

(dy)/dx = [1 + y * e^(2-x) +yx^(-2) * cos(y/x)]/[1 + e^(2-x) - x^(-1)cos(y/x)]dydx=1+ye2x+yx2cos(yx)1+e2xx1cos(yx)

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)])