# How do you implicitly differentiate  sqrt(xy) = x-y/(x-1)?

Jul 27, 2018

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

#### Explanation:

Here ,

$\sqrt{x y} = x - \frac{y}{x - 1}$

Diff. each term w.r.t. $x$, (using product,quotient and chain rules)

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

$\implies \frac{1}{2 \sqrt{x y}} \left[x {y}_{1} + y \cdot 1\right] = 1 - {y}_{1} / \left(x - 1\right) + \frac{y}{x - 1} ^ 2$

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

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

y_1{x/(2sqrt(xy))+1/(x-1)}=(2sqrt(xy)(x-1)^2+2ysqrt(xy)-y(x- 1)^2)/(2sqrt(xy)(x-1)^2)

${y}_{1} \left\{\frac{{x}^{2} - x + 2 \sqrt{x y}}{2 \sqrt{x y} \left(x - 1\right)}\right\}$=(2sqrt(xy)(x- 1)^2+2ysqrt(xy)-y(x-1)^2)/(2sqrt(xy)(x-1)^2)

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

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

Where, ${y}_{1} = \frac{\mathrm{dy}}{\mathrm{dx}}$