Question

निम्नलिखित फलनो का x के सापेक्ष अवकल-गुणांक ज्ञात कीजिए

Find the following calculations of differential calculus relative to x

$\bf \: \frac{1}{ {x}^{n} + {a}^{n} } \:, \: \frac{1}{ \sqrt{x + a} }$

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Answer 1

Answer: $\frac{-nx^{n-1}}{(x^n+a^n)^2}\;\;\textbf{and}\;\;\frac{-1}{2(x+a)\sqrt{x+a}}$

Step-by-step explanation:

First Function:-

$y=\frac{1}{x^n+a^n}\\\;\\y=(x^n+a^n)^{-1}\\\;\\\textbf{Diff. both sides with respect to x}\\\;\\\frac{dy}{dx}=-1(x^n+a^n)^{-1-1}\times\frac{d(x^n+a^n)}{dx}\\\;\\\frac{dy}{dx}=\frac{-1}{(x^n+a^n)^2}\times(nx^{n-1}+0)\\\;\\\frac{dy}{dx}=\frac{-nx^{n-1}}{(x^n+a^n)^2}$

Second Function:-

$y=\frac{1}{\sqrt{x+a}}\\\;\\y=(x+a)^{-\frac{1}{2}}\\\;\\\textbf{Diff. both sides with respect to x}\\\;\\\frac{dy}{dx}=-\frac{1}{2}(x+a)^{-\frac{1}{2}-1}\times\frac{d(x+a)}{dx}\\\;\\\frac{dy}{dx}=-\frac{1}{2}(x+a)^{\frac{-1-2}{2}}\times(1+0)\\\;\\\frac{dy}{dx}=-\frac{1}{2}(x+a)^{-\frac{3}{2}}\\\;\\\frac{dy}{dx}=\frac{-1}{2(x+a)\sqrt{x+a}}$

Answer 2

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