# How do you implicitly differentiate  2x/y = ysqrt(x^2+y^2)-x?

Apr 24, 2018

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

#### Explanation:

Alright, this is a very long one. I'll number each step to make it easier, and also I didn't combine steps so you knew what was going on.

$2 x {y}^{-} 1 = y {\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}} - x$

First we take $\frac{d}{\mathrm{dx}}$ of each term:
2. $\frac{d}{\mathrm{dx}} \left[2 x {y}^{-} 1\right] = \frac{d}{\mathrm{dx}} \left[y {\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}}\right] - \frac{d}{\mathrm{dx}} \left[x\right]$
3. $\frac{d}{\mathrm{dx}} \left[2 x\right] {y}^{-} 1 + x \frac{d}{\mathrm{dx}} \left[{y}^{-} 1\right] = \frac{d}{\mathrm{dx}} \left[y\right] {\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}} + y \frac{d}{\mathrm{dx}} \left[{\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}}\right] - \frac{d}{\mathrm{dx}} \left[x\right]$
4. $2 {y}^{-} 1 + x \frac{d}{\mathrm{dx}} \left[{y}^{-} 1\right] = \frac{d}{\mathrm{dx}} \left[y\right] {\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}} + \frac{y {\left({x}^{2} + {y}^{2}\right)}^{- \frac{1}{2}}}{2} \frac{d}{\mathrm{dx}} \left[{x}^{2} + {y}^{2}\right] - 1$
5. $2 {y}^{-} 1 + x \frac{d}{\mathrm{dx}} \left[{y}^{-} 1\right] = \frac{d}{\mathrm{dx}} \left[y\right] {\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}} + \frac{y {\left({x}^{2} + {y}^{2}\right)}^{- \frac{1}{2}}}{2} \left(\frac{d}{\mathrm{dx}} \left[{x}^{2}\right] + \frac{d}{\mathrm{dx}} \left[{y}^{2}\right]\right) - 1$
6. $2 {y}^{-} 1 + x \frac{d}{\mathrm{dx}} \left[{y}^{-} 1\right] = \frac{d}{\mathrm{dx}} \left[y\right] {\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}} + \frac{y {\left({x}^{2} + {y}^{2}\right)}^{- \frac{1}{2}}}{2} \left(2 x + \frac{d}{\mathrm{dx}} \left[{y}^{2}\right]\right) - 1$

Now we use $\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$:
7. $2 {y}^{-} 1 - \frac{\mathrm{dy}}{\mathrm{dx}} x {y}^{-} 2 = \frac{\mathrm{dy}}{\mathrm{dx}} {\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}} + \frac{y {\left({x}^{2} + {y}^{2}\right)}^{- \frac{1}{2}}}{2} \left(2 x + \frac{\mathrm{dy}}{\mathrm{dx}} 2 y\right) - 1$
8. Now we rearrange:
$- \frac{\mathrm{dy}}{\mathrm{dx}} \left(x {y}^{-} 2 - {\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}}\right) = y x {\left({x}^{2} + {y}^{2}\right)}^{- \frac{1}{2}} + \frac{\mathrm{dy}}{\mathrm{dx}} {y}^{2} {\left({x}^{2} + {y}^{2}\right)}^{- \frac{1}{2}} - 1 - 2 {y}^{-} 1$
9. $- \frac{\mathrm{dy}}{\mathrm{dx}} \left(x {y}^{-} 2 - {\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}} + {y}^{2} {\left({x}^{2} + {y}^{2}\right)}^{- \frac{1}{2}}\right) = y x {\left({x}^{2} + {y}^{2}\right)}^{- \frac{1}{2}} - 1 - 2 {y}^{-} 1$
10. $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y x {\left({x}^{2} + {y}^{2}\right)}^{- \frac{1}{2}} - 1 - 2 {y}^{-} 1}{x {y}^{-} 2 - {\left({x}^{2} + {y}^{2}\right)}^{\frac{1}{2}} + {y}^{2} {\left({x}^{2} + {y}^{2}\right)}^{- \frac{1}{2}}}$