Question

Differentiate the function w.r.t.x. : $\frac{e^{ \sqrt{x}}+1}{e^{ \sqrt{x}}-1}$

Answer 1

another good question.

Answer 2

Answer:

$\frac{d}{dx} (\frac{e^{\sqrt{x} } +1}{e^{\sqrt{x} }-1 } )= -\frac{e^{\sqrt{x} } }{\sqrt{x}(e^{\sqrt{x} }-1) ^{2} }$

Step-by-step explanation:

Formula used

$\frac{d}{dx} (\frac{U}{V} )=\frac{V.\frac{dU}{dx} -U.\frac{dV}{dx} }{V^{2} }$

so here

$U =e^{\sqrt{x} }+1\\ \\V=e^{\sqrt{x} }-1$

$\frac{d}{dx} (\frac{e^{\sqrt{x} } +1}{e^{\sqrt{x} }-1 } )=\frac{(e^{\sqrt{x} }-1 ).\frac{d(e^{\sqrt{x} }+1 )}{dx} -(e^{\sqrt{x} }+1) .\frac{d(e^{\sqrt{x} }-1) }{dx} }{(e^{\sqrt{x} }-1) ^{2} }\\ \\ \\ \\= \frac{(e^{\sqrt{x} }-1 ).\frac{(e^{\sqrt{x} } )}{2\sqrt{x} } -(e^{\sqrt{x} }+1) .\frac{(e^{\sqrt{x} }) }{2\sqrt{x} } }{(e^{\sqrt{x} }-1) ^{2} }$

$=\frac{\frac{(e^{\sqrt{x} } )}{2\sqrt{x} }[ (e^{\sqrt{x} }-1 )-(e^{\sqrt{x} }+1) ] }{(e^{\sqrt{x} }-1) ^{2} }\\ \\ \\ \\=\frac{\frac{(e^{\sqrt{x} } )}{2\sqrt{x} }[ (e^{\sqrt{x} }-1 -e^{\sqrt{x} }-1) ] }{(e^{\sqrt{x} }-1) ^{2} }\\ \\ \\ \\ =\frac{\frac{(e^{\sqrt{x} } )}{2\sqrt{x} }[-2] }{(e^{\sqrt{x} }-1) ^{2} }\\ \\ \\ \\=-\frac{e^{\sqrt{x} } }{\sqrt{x}(e^{\sqrt{x} }-1) ^{2} }$