If y = c{e^{{{sin }^{ - 1}}x}},then the corre | vedclass

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Question

(A) Given the equation: $y = c{e^{{{\sin }^{ - 1}}x}}$.Differentiating both sides with respect to $x$ using the chain rule:$\frac{dy}{dx} = c{e^{{{\si

Vedclass · Accepted Answer

$\frac{dy}{dx} = \frac{y}{\sqrt{1 - x^2}}$

Vedclass · Answer

(A) Given the equation: $y = c{e^{{{\sin }^{ - 1}}x}}$.Differentiating both sides with respect to $x$ using the chain rule:$\frac{dy}{dx} = c{e^{{{\sin }^{ - 1}}x}} \cdot \frac{d}{dx}(\sin^{-1}x)$We know that $\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1 - x^2}}$.Substituting this into the derivative:$\frac{dy}{dx} = c{e^{{{\sin }^{ - 1}}x}} \cdot \frac{1}{\sqrt{1 - x^2}}$Since $y = c{e^{{{\sin }^{ - 1}}x}}$,we can substitute $y$ back into the equation:$\frac{dy}{dx} = \frac{y}{\sqrt{1 - x^2}}$Thus,the correct option is $A$.

If $y = c{e^{{{\sin }^{ - 1}}x}}$,then the corresponding differential equation is:

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