Find frac{d^{2} y}{d x^{2}}, if y=x^{3}+tan x | vedclass

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(A) Given that $y = x^{3} + 	an x$. First,we find the first derivative with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^{3}) + \frac{d}{dx}(	an

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$6 x + 2 \sec^{2} x 	an x$

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(A) Given that $y = x^{3} + 	an x$. First,we find the first derivative with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^{3}) + \frac{d}{dx}(	an x) = 3x^{2} + \sec^{2} x$. Now,we find the second derivative by differentiating $\frac{dy}{dx}$ with respect to $x$: $\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(3x^{2} + \sec^{2} x) = \frac{d}{dx}(3x^{2}) + \frac{d}{dx}(\sec^{2} x)$. Using the power rule and chain rule: $\frac{d}{dx}(3x^{2}) = 6x$. For $\frac{d}{dx}(\sec^{2} x)$,let $u = \sec x$,then $\frac{d}{dx}(u^{2}) = 2u \frac{du}{dx} = 2(\sec x)(\sec x 	an x) = 2 \sec^{2} x 	an x$. Thus,$\frac{d^{2}y}{dx^{2}} = 6x + 2 \sec^{2} x 	an x$.

Find $\frac{d^{2} y}{d x^{2}},$ if $y=x^{3}+\tan x$.

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