Question

\sqrt{e}^{\sqrt{x}} x > 0 dy/dx ज्ञात कीजिए

Answer 1

Question :-

$\sf \: if \: y \: = \sqrt{ {e}^{ \sqrt{x} } } \: \: then \: find \: \frac{dy}{dx}$

Answer :-

$\boxed{ \to \sf \frac{dy}{dx} = \frac{ {e}^{ \sqrt{ x } } }{4 \sqrt{x {e}^{ \sqrt{x} } } } } \:$

Solution :-

We have ,

$\to \sf y = \sqrt{ {e}^{ \sqrt{x} } } \\ \\ \sf \red{ differentiate \: with \: respect \: to \: x} \\ \\ \to \sf \frac{dy}{dx} = \frac{d}{dx} \sqrt{ {e}^{ \sqrt{x} } } \\ \\ \boxed{ \because \sf \frac{d}{dx} \sqrt{x } = { \frac{1}{2 \sqrt{x} } } } \\ \\ \to \sf \frac{dy}{dx} = \frac{1}{2 \sqrt{ {e}^{ \sqrt{x} } } } \: \frac{d}{dx} {e}^{ \sqrt{x} } \\ \\ \boxed{ \because \sf \frac{d}{dx} {e}^{x} = {e}^{x} } \\ \therefore \\ \\ \to \sf \frac{dy}{dx} = \frac{1}{2 \sqrt{ {e}^{ \sqrt{x} } } } {e}^{ \sqrt{x} } \: \frac{d}{dx} \sqrt{x} \\ \\ \to \sf \frac{dy}{dx} = \frac{ {e}^{ \sqrt{x} } }{2 \sqrt{ {e}^{ \sqrt{x} } } } \times \frac{1}{2 \sqrt{x} } \\ \\ \sf \frac{dy}{dx} = \frac{ {e}^{ \sqrt{x} } }{4 \sqrt{x} \sqrt{ {e}^{ \sqrt{x} } } } \\ \\ \boxed{ \to \sf \frac{dy}{dx} = \frac{ {e}^{ \sqrt{ x } } }{4 \sqrt{x {e}^{ \sqrt{x} } } } }$

Answer 2

$dy/dx = \frac{{e^{\sqrt{x}} }}{4\sqrt{xe^{\sqrt{x}} }}$  

Step-by-step explanation:

dy/dx ज्ञात कीजिए

$y = \sqrt{e^{\sqrt{x}} }$

$dy/dx = d(\sqrt{e^{\sqrt{x}} })/dx$

=> $dy/dx = \frac{1}{2\sqrt{e^{\sqrt{x}} }} \times d({e^{\sqrt{x}} })/dx$

=>  $dy/dx = \frac{1}{2\sqrt{e^{\sqrt{x}} }} \times ({e^{\sqrt{x}} }) \times d({\sqrt{x}} })/dx$

=> $dy/dx = \frac{1}{2\sqrt{e^{\sqrt{x}} }} \times ({e^{\sqrt{x}} }) \times \frac{1}{2\sqrt{x} }$

=> $dy/dx = \frac{{e^{\sqrt{x}} }}{4\sqrt{x} \sqrt{e^{\sqrt{x}} }}$

=> $dy/dx = \frac{{e^{\sqrt{x}} }}{4\sqrt{xe^{\sqrt{x}} }}$

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