# (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)?

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

$\sin \left(\frac{y}{x}\right) + y {e}^{2 - x} = x - y$

To get $\frac{\mathrm{dy}}{\mathrm{dx}}$, you need to use implicit differentiation. Differentiate both sides with respect to $x$

$\frac{d}{\mathrm{dx}} \left[\sin \left(\frac{y}{x}\right) + y {e}^{2 - x}\right] = \frac{d}{\mathrm{dx}} \left(x - y\right)$

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

$\frac{d}{\mathrm{dx}} \left[\sin \left(\frac{y}{x}\right)\right] = \cos \left(\frac{y}{x}\right) \cdot \frac{d}{\mathrm{dx}} \left(y \cdot {x}^{- 1}\right)$

$\frac{d}{\mathrm{dx}} \left[\sin \left(\frac{y}{x}\right)\right] = \cos \left(\frac{y}{x}\right) \cdot \left[\frac{\mathrm{dy}}{\mathrm{dx}} \cdot {x}^{- 1} + y \cdot {\left(- x\right)}^{- 2}\right]$

Next

$\frac{d}{\mathrm{dx}} \left(y {e}^{2 - x}\right) = \frac{\mathrm{dy}}{\mathrm{dx}} \cdot {e}^{2 - x} + y \cdot \frac{d}{\mathrm{dx}} \left({e}^{2 - x}\right)$

$\frac{d}{\mathrm{dx}} \left(y {e}^{2 - x}\right) = \frac{\mathrm{dy}}{\mathrm{dx}} \cdot {e}^{2 - x} + y \cdot {e}^{2 - x} \cdot \left(- 1\right)$

$\frac{d}{\mathrm{dx}} \left(y {e}^{2 - x}\right) = \frac{\mathrm{dy}}{\mathrm{dx}} \cdot {e}^{2 - x} - y \cdot {e}^{2 - x}$

Plug these back into your target calculation to get

$\cos \left(\frac{y}{x}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \cdot {x}^{- 1} + \cos \left(\frac{y}{x}\right) \cdot \left(- y {x}^{- 2}\right) + \frac{\mathrm{dy}}{\mathrm{dx}} \cdot {e}^{2 - x} - y \cdot {e}^{2 - x} = 1 - \frac{\mathrm{dy}}{\mathrm{dx}}$

Isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$ on one side of the equation to get

$\cos \left(\frac{y}{x}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \cdot {x}^{- 1} + \frac{\mathrm{dy}}{\mathrm{dx}} \cdot {e}^{2 - x} + \left(\mathrm{dy}\right) = 1 + y \cdot {e}^{2 - x} + y {x}^{- 2} \cdot \cos \left(\frac{y}{x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \left[{x}^{- 1} \cos \left(\frac{y}{x}\right) + {e}^{2 - x} + 1\right] = 1 + y \cdot {e}^{2 - x} + y {x}^{- 2} \cdot \cos \left(\frac{y}{x}\right)$

Finally, you have

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 + y \cdot {e}^{2 - x} + y {x}^{- 2} \cdot \cos \left(\frac{y}{x}\right)}{1 + {e}^{2 - x} - {x}^{- 1} \cos \left(\frac{y}{x}\right)}$

You can simplify this to the form

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} + y \cdot {x}^{2} \cdot {e}^{2 - x} + y \cos \left(\frac{y}{x}\right)}{x} ^ \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}{x + x \cdot {e}^{2 - x} - \cos \left(\frac{y}{x}\right)}$

(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

$x {y}^{2} - \sin \left(x + 2 y\right) = 2 x$

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

$\frac{d}{\mathrm{dx}} \left[x {y}^{2} - \sin \left(x + 2 y\right)\right] = \frac{d}{\mathrm{dx}} \left(2 x\right)$

This will get you

$\frac{d}{\mathrm{dx}} \left(x {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(x\right) \cdot {y}^{2} + x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \cdot 2 y$

$\frac{d}{\mathrm{dx}} \left(x {y}^{2}\right) = {y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}}$

and

$\frac{d}{\mathrm{dx}} \left[\sin \left(x + 2 y\right)\right] = \cos \left(x + 2 y\right) \cdot \frac{d}{\mathrm{dx}} \left(x + 2 y\right)$

$\frac{d}{\mathrm{dx}} \left[\sin \left(x + 2 y\right)\right] = \cos \left(x + 2 y\right) \cdot \left[\frac{d}{\mathrm{dx}} \left(x\right) + 2 \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

$\frac{d}{\mathrm{dx}} \left[\sin \left(x + 2 y\right)\right] = \cos \left(x + 2 y\right) \cdot \left(1 + 2 \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Plug these back into your target calculation to get

${y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - \left[\cos \left(x + 2 y\right) + 2 \cos \left(x + 2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}}\right] = 2$

${y}^{2} + 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - \cos \left(x + 2 y\right) - 2 \cos \left(x + 2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 2$

Once again, isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$ on one side

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot 2 \left[x y - \cos \left(x + 2 y\right)\right] = 2 - {y}^{2} + \cos \left(x + 2 y\right)$

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