Question

निम्नलिखित फलनों के अवकलन ज्ञात कीजिए (यह समझा जाय कि a, b, c, d,p, q, r और s निशिचत शून्येतर अचर हैं m तथा n पूर्णांक हैंl ) : $\dfrac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x}}$

Answer 1

Answer:

Step-by-step explanation:

$\dfrac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x}}$

माना  कि  $y=\dfrac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x}}$

∴ अवकलन करने पर -

$\dfrac{dy}{dx} =\dfrac{d}{dx} [\dfrac{1 + \dfrac{1}{x}}{1 - \dfrac{1}{x}}]\\\\\dfrac{dy}{dx}=\dfrac{(1-\dfrac{1}{x})\dfrac{d}{dx}(1+\dfrac{1}{x} )-(1+\dfrac{1}{x} )\dfrac{d}{dx} (1-\dfrac{1}{x} )}{(1-\dfrac{1}{x} )^2}\\\\\dfrac{dy}{dx}=\dfrac{(1-\dfrac{1}{x}) (-\dfrac{1}{x^2})-(1+\dfrac{1}{x})(\dfrac{1}{x^2})    }{(1-\dfrac{1}{x} )^2}\\ \\\dfrac{dy}{dx}=\dfrac{-\dfrac{1}{x^2}+\dfrac{1}{x^3}-\dfrac{1}{x^2}-\dfrac{1}{x^3}}{(1-\dfrac{1}{x})^2}\\\\\dfrac{dy}{dx}=\dfrac{-2}{x^2(1-\dfrac{1}{x})^2}$

$\dfrac{dy}{dx}=\dfrac{-2}{x^2(\dfrac{x-1}{x})^2}\\\\\dfrac{dy}{dx}=\dfrac{-2x^}{x^2(x-1)^2}\\\\\dfrac{dy}{dx}=\dfrac{-2}{(x-1)^2}$