Question

If y=sin^(-1)(x) then (dy/dx) ​

Answer 1

Answer

• dy/dx = 1/√(1-x²)

Given

• y = sin⁻¹x

To Find

• dy/dx

Solution

y = sin⁻¹x

Differentiating with respect to " x " on both sides , we get ,

$\implies \bf \dfrac{dy}{dx}=\dfrac{d}{dx}(sin^{-1}x)\\\\\implies \bf \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}$

More Info

$\bullet \bf \;\; \dfrac{d}{dx}(sin^{-1}x)=\dfrac{1}{\sqrt{1-x^2}}\\\\\bullet \bf \;\; \dfrac{d}{dx}(cos^{-1}x)=\dfrac{-1}{\sqrt{1-x^2}}\\\\\bullet \bf \;\; \dfrac{d}{dx}(tan^{-1}x)=\dfrac{1}{1+x^2}\\\\\bullet \bf \;\; \dfrac{d}{dx}(cot^{-1}x)=\dfrac{-1}{1+x^2}\\\\\bullet \bf \;\; \dfrac{d}{dx}(sec^{-1}x)=\dfrac{1}{\mid x \mid \sqrt{x^2-1}}\\\\\bullet \bf \;\; \dfrac{d}{dx}(cosec^{-1}x)=\dfrac{-1}{\mid x \mid \sqrt{x^2-1}}$

Answer 2

Given ,

The function is $\rm y = {Sin}^{ - 1}( x)$

It can be written as ,

Sin(y) = x

Differentiating with respect to x , we get

$\sf \mapsto \frac{d(x)}{dx} = \frac{dSin(y)}{dx} \\ \\ \sf \mapsto 1 = Cos(y) \times \frac{d(y)}{dx} \\ \\ \sf \mapsto \frac{dy}{dx} = \frac{1}{Cos(y)} \\ \\ \sf \mapsto \frac{dy}{dx} = \frac{1}{ \sqrt{1 - {Sin}^{2}y } } \\ \\ \sf \mapsto \frac{dy}{dx} = \frac{1}{ \sqrt{1 - { \{Sin ({Sin}^{ - 1} (x) )\}}^{2} } } \: \: \: \: \{ \because \: y = {Sin}^{ - 1}( x) \: \}\\ \\ \sf \mapsto \frac{dy}{dx} = \frac{1}{ \sqrt{1 - {(x)}^{2} } }$

Remmember :

$\sf \mapsto \frac{d \{Sin(x) \} }{dx} = Cos(x) \\ \\ \sf \mapsto {Sin}^{2} (x) + {Cos}^{2} (x) = 1$