Question

$please \: give \: derivative \: of \: picture \: above$
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Answer 1

Answer:

$\tt \dfrac{dy}{dx} =\dfrac{-2a^2x}{(a^2-x^2)^{1/2}\times (a^2+x^2)^{3/2}}$

Explanation:

Given:

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

To Find:

dy/dx

Solution:

Here let,

$\tt t=\dfrac{a^2-x^2}{a^2+x^2}$

Now differentiate on both sides with respect to x,

$\tt \dfrac{dt}{dx} =\dfrac{d}{dx}\bigg( \dfrac{a^2-x^2}{a^2+x^2}\bigg)$

By quotient rule we know that,

$\tt \bigg(\dfrac{u}{v}\bigg)'=\dfrac{u'v-v'u}{v^2}$

Therefore,

$\tt \dfrac{dt}{dx} =\dfrac{(a^2+x^2)(-2x)-(a^2-x^2)(2x)}{(a^2+x^2)^2}$

$\tt \implies \dfrac{-2a^2x-2x^3-(2a^2x-2x^3)}{(a^2+x^2)^2}$

$\tt \implies \dfrac{-2a^2x-2x^3-2a^2x+2x^3}{(a^2+x^2)^2}$

$\tt \implies \dfrac{-4a^2x}{(a^2+x^2)^2}$

Now if,

$\tt t=\dfrac{a^2-x^2}{a^2+x^2},$

then y = √t

Differentiate on both sides with respect to t,

$\tt \dfrac{dy}{dt} =\dfrac{1}{2\sqrt{t} }$

Now,

$\tt \dfrac{dy}{dx} =\dfrac{dy}{dt} .\dfrac{dt}{dx}$

Substitute the values,

$\tt \dfrac{dy}{dx} =\dfrac{1}{2\sqrt{t} } \times \dfrac{-4a^2x}{(a^2+x^2)^2}$

$\tt \implies \dfrac{-2a^2x}{\sqrt{t}\times (a^2+x^2)^2 }$

$\tt \implies \dfrac{-2a^2x}{\sqrt{\dfrac{a^2-x^2}{a^2+x^2} }\times (a^2+x^2)^2 }$

$\tt \implies \dfrac{-2a^2x\sqrt{a^2+x^2} }{\sqrt{a^2-x^2} \times (a^2+x^2)^2}$

$\tt \implies \dfrac{-2a^2x(a^2+x^2)^{1/2}}{(a^2-x^2)^{1/2}\times (a^2+x^2)^2}$

$\tt \implies \dfrac{-2a^2x}{(a^2-x^2)^{1/2}\times (a^2+x^2)^{3/2}}$