If y=e^{tan ^{-1} x}, then: | vedclass

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(A) Given $y=e^{	an ^{-1} x}$.Differentiating with respect to $x$, we get:$\frac{dy}{dx} = e^{	an ^{-1} x} 	imes \frac{d}{dx}(	an ^{-1} x) = e^{

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$(1+x^2) y_2+(2 x-1) y_1=0$

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(A) Given $y=e^{	an ^{-1} x}$.Differentiating with respect to $x$, we get:$\frac{dy}{dx} = e^{	an ^{-1} x} 	imes \frac{d}{dx}(	an ^{-1} x) = e^{	an ^{-1} x} 	imes \frac{1}{1+x^2}$.Since $y = e^{	an ^{-1} x}$, we can write:$\frac{dy}{dx} = \frac{y}{1+x^2}$.Multiplying both sides by $(1+x^2)$, we get:$(1+x^2) y_1 = y$.Differentiating again with respect to $x$ using the product rule on the left side:$\frac{d}{dx}[(1+x^2) y_1] = \frac{d}{dx}(y)$.$(1+x^2) y_2 + y_1(2x) = y_1$.Rearranging the terms:$(1+x^2) y_2 + (2x - 1) y_1 = 0$.

If $y=e^{\tan ^{-1} x}$, then:

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