Question

If$\sqrt{x} +\sqrt{y} =\sqrt{a}$ find second order derivative
at x=a

Answer 1

Answer:

$\large\boxed{\sf{-\infty}}$

Step-by-step explanation:

Given an equation,

$\sqrt{x} + \sqrt{y} = \sqrt{a}$

Differentiating both the sides wrt x, we get,

$= > \dfrac{d}{dx} \sqrt{x} + \dfrac{d}{dx} \sqrt{y} = \dfrac{d}{dx} \sqrt{a}$

But, we know that,

• $\dfrac{d}{dx} {x}^{n} = n {x}^{n - 1}$

• $\dfrac{d}{dx} c = 0$

where, c is any constant term.

Therefore, we will get,

$= > \dfrac{1}{2 \sqrt{x} } + \dfrac{1}{2 \sqrt{y} } \dfrac{dy}{dx} = 0 \\ \\ = > \frac{1}{ \sqrt{x} } + \dfrac{1}{ \sqrt{y} } \dfrac{dy}{dx} = 0 \\ \\ = > \dfrac{1}{ \sqrt{y} } \dfrac{dy}{dx} = - \dfrac{1}{ \sqrt{x} } \\ \\ = > \dfrac{dy}{dx} = - \dfrac{ \sqrt{y} }{ \sqrt{x} }$

Again differentiatiating both sides, wrt x,

$= > \dfrac{ {d}^{2}y }{d {x}^{2} } = - \dfrac{ \sqrt{x} \frac{d}{dx} \sqrt{y} - \sqrt{y} \frac{d}{dx} \sqrt{x} }{ {( \sqrt{x} )}^{2} } \\ \\ = > \dfrac{ {d}^{2}y }{d {x}^{2} } = - \dfrac{ \frac{ \sqrt{x} }{2 \sqrt{y} } - \frac{ \sqrt{y} }{2 \sqrt{x} } }{x} \\ \\ = > \dfrac{ {d}^{2} y}{d {x}^{2} } = -\dfrac{x - y}{2x \sqrt{xy} }$

Now, when x = a,

$= > \sqrt{a} + \sqrt{y} = \sqrt{a} \\ \\ = > \sqrt{y } = 0 \\ \\ = > y = 0$

Therefore, second order derivative at x = a is,

$=- \dfrac{a - 0}{2a \sqrt{a \times 0} } \\ \\ = - \frac{a}{0} \\ \\ = - \infty$

Hence, the required value of second order derivative at x = a , is $\large\bold{-\infty}$.