Question

derivative of
$\sqrt{a {} ^ {2} } - \sqrt{x {}^{2} } \div \sqrt{a { }^{2} } + \sqrt{x {}^{2} }$
​

Answer 1

Given :

$\bullet\ \; \sf \dfrac{\sqrt{a^2}-\sqrt{x^2}}{\sqrt{a^2}+\sqrt{x^2}}$

To Find :

Derivative of the given value

Solution :

Let ,

$\sf y=\dfrac{\sqrt{a^2}-\sqrt{x^2}}{\sqrt{a^2}+\sqrt{x^2}}\\\\\to \sf y=\dfrac{a-x}{a+x}$

Differentiate with respect to x on both sides ,

$\to \sf \dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\dfrac{a-x}{a+x}\bigg)$

• d/dx(u/v) = (vu' - uv') / v²
• u' ➠ du/dx
• v' ➠ dv/dx

$\to \sf \dfrac{dy}{dx}=\dfrac{(a+x)\frac{d}{dx}(a-x)-(a-x)\frac{d}{dx}(a+x)}{(a+x)^2}$

$\to \sf \dfrac{dy}{dx}=\dfrac{(a+x)(-1)-(a-x)(1)}{(a+x)^2}$

$\to \sf \dfrac{dy}{dx}=\dfrac{-a-x-a+x}{(a+x)^2}$

$\leadsto \sf \pink{\dfrac{dy}{dx}=\dfrac{-2a}{(a+x)^2}}\ \; \bigstar$

Answer 2

Answer:

Given :

\bullet\ \; \sf \dfrac{\sqrt{a^2}-\sqrt{x^2}}{\sqrt{a^2}+\sqrt{x^2}}∙

a

2

+

x

2

a

2

−

x

2

To Find :

Derivative of the given value

Solution :

Let ,

\begin{gathered}\sf y=\dfrac{\sqrt{a^2}-\sqrt{x^2}}{\sqrt{a^2}+\sqrt{x^2}}\\\\\to \sf y=\dfrac{a-x}{a+x}\end{gathered}

y=

a

2

+

x

2

a

2

−

x

2

→y=

a+x

a−x

Differentiate with respect to x on both sides ,

\to \sf \dfrac{dy}{dx}=\dfrac{d}{dx}\bigg(\dfrac{a-x}{a+x}\bigg)→

dx

dy

=

dx

d

(

a+x

a−x

)

d/dx(u/v) = (vu' - uv') / v²

u' ➠ du/dx

v' ➠ dv/dx

\to \sf \dfrac{dy}{dx}=\dfrac{(a+x)\frac{d}{dx}(a-x)-(a-x)\frac{d}{dx}(a+x)}{(a+x)^2}→

dx

dy

=

(a+x)

2

(a+x)

dx

d

(a−x)−(a−x)

dx

d

(a+x)

\to \sf \dfrac{dy}{dx}=\dfrac{(a+x)(-1)-(a-x)(1)}{(a+x)^2}→

dx

dy

=

(a+x)

2

(a+x)(−1)−(a−x)(1)

\to \sf \dfrac{dy}{dx}=\dfrac{-a-x-a+x}{(a+x)^2}→

dx

dy

=

(a+x)

2

−a−x−a+x

\leadsto \sf \pink{\dfrac{dy}{dx}=\dfrac{-2a}{(a+x)^2}}\ \; \bigstar⇝

dx

dy

=

(a+x)

2

−2a

★