frac{d}{dx} left( e^{sqrt{1 - x^2}} cdot tan | vedclass

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(B) To find the derivative of $f(x) = e^{\sqrt{1 - x^2}} \cdot 	an x$,we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$

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$e^{\sqrt{1 - x^2}} \left[ \sec^2 x - \frac{x 	an x}{\sqrt{1 - x^2}} ight]$

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(B) To find the derivative of $f(x) = e^{\sqrt{1 - x^2}} \cdot 	an x$,we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.Let $u = e^{\sqrt{1 - x^2}}$ and $v = 	an x$.Then $\frac{dv}{dx} = \sec^2 x$.Using the chain rule for $u$,$\frac{du}{dx} = e^{\sqrt{1 - x^2}} \cdot \frac{d}{dx}(\sqrt{1 - x^2}) = e^{\sqrt{1 - x^2}} \cdot \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = -\frac{x e^{\sqrt{1 - x^2}}}{\sqrt{1 - x^2}}$.Applying the product rule:$\frac{d}{dx} (u v) = e^{\sqrt{1 - x^2}} \cdot \sec^2 x + 	an x \cdot \left( -\frac{x e^{\sqrt{1 - x^2}}}{\sqrt{1 - x^2}} ight)$.Factoring out $e^{\sqrt{1 - x^2}}$,we get:$= e^{\sqrt{1 - x^2}} \left[ \sec^2 x - \frac{x 	an x}{\sqrt{1 - x^2}} ight]$.

$\frac{d}{dx} \left( e^{\sqrt{1 - x^2}} \cdot \tan x \right)$

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