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

May 20, 2018

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

Explanation:

start by differentiating both sides w.r.t x

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

lets look at this term by term,
RHS,
the derivative of xy with respect to x using the product rule comes as $\frac{d}{\mathrm{dx}} \left(x\right) \cdot y + x \cdot \frac{d}{\mathrm{dx}} \left(y\right)$ which is equal to $y + x \frac{\mathrm{dy}}{\mathrm{dx}}$

we need to use the chain rule and the product rule for
differentiating $2 {e}^{y} \sqrt{x - y}$

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

the derivative of $2 {e}^{y}$ w.r.t x is $2 {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}$

and the derivative of $\sqrt{x - y} = {\left(x - y\right)}^{\frac{1}{2}}$ using the chain rule comes as $\frac{1}{2 \sqrt{x - y}} \cdot \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

therefore the derivative of the second term is
$2 {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} \cdot \sqrt{x - y} + \frac{2 {e}^{y}}{2 \sqrt{x - y}} \cdot \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

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

therefore, the entire equation becomes

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

multiply both the sides by -1 to make the left hand side positive,

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

Now shift all the dy/dx terms to the left side and factorise

dy/dx + x dy/dx + 2e^y sqrt(x-y)dy/dx - e^y/(sqrt(x-y)) dy/dx = -y - e^y/(sqrt(x-y))

dy/dx (1 + x + 2e^y sqrt(x-y) - e^y/(sqrt(x-y))) = -y - e^y/(sqrt(x-y)

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

May 20, 2018

The answer is $= - \frac{y + {e}^{y} \cdot \frac{1}{\sqrt{x - y}}}{1 + x + 2 {e}^{y} \sqrt{x - y} - {e}^{y} / \left(\sqrt{x - y}\right)}$

Explanation:

Let,

$f \left(x , y\right) = y + x y + 2 {e}^{y} \sqrt{x - y}$

And

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$

$\frac{\partial f}{\partial x} = y + 2 {e}^{y} \cdot \frac{1}{2 \sqrt{x - y}}$

$\frac{\partial f}{\partial y} = 1 + x + 2 {e}^{y} \sqrt{x - y} - {e}^{y} / \left(\sqrt{x - y}\right)$

Therefore,

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