Question

y = e^{sin^{-1} x} dy/dx ज्ञात कीजिए

Answer 1

Question :-

$\sf \: if \: y = {e}^{ ({ \sin}^{ - 1}x) } \: then \: find \: \frac{dy}{dx}$

Answer :-

$\to \boxed{ \sf \frac{dy}{dx} = \frac{ {e}^{ ({ \sin}^{ - 1}x) }}{ \sqrt{1 - {x}^{2} } } } \:$

Solution :-

We have ,

$\mapsto\sf \: y = {e}^{ ({ \sin}^{ - 1}x) } \: \\ \\ \bf \red{ differentiate \: with \: respect \: to \: x} \\ \\ \mapsto \sf \: \frac{dy}{dx} = \frac{d}{dx} {e}^{ ({ \sin}^{ - 1}x) } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \sf \frac{d}{dx} {e}^{x} = {e}^{x} } \\ \to \sf \frac{dy}{dx} = {e}^{ ({ \sin}^{ - 1}x) } \frac{d}{dx} ({ \sin}^{ - 1}x) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\because \sf\frac{d}{dx} ({ \sin}^{ - 1}x) = \frac{1}{ \sqrt{1 - {x}^{2} } } } \\ \\ \to \boxed{ \sf \frac{dy}{dx} = \frac{ {e}^{ ({ \sin}^{ - 1}x) }}{ \sqrt{1 - {x}^{2} } } }$

Some same questions for practice :-

(1) if y = sin{cos(x)} then find its derivative .

(2) if xy = x²y + yx+ xy³ ,find $\: \frac{dy}{dx}$

Answer 2

dy/dx   = e^(Sin⁻¹x)  / √(1 - x²) यदि y  =  e^(Sin⁻¹x)

Step-by-step explanation:

y  =  e^(Sin⁻¹x)

dy/dx   =     e^(Sin⁻¹x)  * d (Sin⁻¹x)/dx

d (Sin⁻¹x)/dx  = 1/√(1 - x²)

=> dy/dx   =     e^(Sin⁻¹x)  *   1/√(1 - x²)

=> dy/dx   = e^(Sin⁻¹x)  / √(1 - x²)

और अधिक जानें :

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