# If y=tan^(-1)(x/y) then dy/dx will be? (a) y/x (b) -y/x (c) x/y (d) -x/y (e) -xy

Jul 27, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{{y}^{2} + {x}^{2} + x}$

#### Explanation:

As $y = {\tan}^{- 1} \left(\frac{x}{y}\right)$, we have $\tan y = \frac{x}{y}$

and differentiating implicitly

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{y}$

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

Now as ${\sec}^{2} y = 1 + {\tan}^{2} y = 1 + {x}^{2} / {y}^{2}$

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

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

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{{y}^{2} + {x}^{2} + x}$

Jul 28, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{{y}^{2} + {x}^{2} + x} ,$

#### Explanation:

$y = {\tan}^{-} 1 \left(\frac{x}{y}\right) \Rightarrow \tan y = \frac{x}{y} , \mathmr{and} , x = y \tan y .$

$\therefore \frac{d}{\mathrm{dy}} \left\{x\right\} = \frac{d}{\mathrm{dy}} \left\{y \tan y\right\} .$

$\therefore \frac{\mathrm{dx}}{\mathrm{dy}} = y \frac{d}{\mathrm{dy}} \left(\tan y\right) + \tan y \cdot \frac{d}{\mathrm{dy}} \left(y\right) , \ldots \ldots . . \text{[Product Rule]}$

$= y {\sec}^{2} y + \tan y = y \left(1 + {\tan}^{2} y\right) + \tan y .$

Subst.ing, $\tan y = \frac{x}{y} ,$ in this, we have,

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{{y}^{2} + {x}^{2} + x} ,$ as Respected Shwetank Mauria has