If y = frac{e^{sin x} + sinh^3 x}{cosh x - ta | vedclass

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(D) Given $y = \frac{e^{\sin x} + \sinh^3 x}{\cosh x - 	an x}$.Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} ight) = \frac{u'v - uv'}{v^

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$2$

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(D) Given $y = \frac{e^{\sin x} + \sinh^3 x}{\cosh x - 	an x}$.Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} ight) = \frac{u'v - uv'}{v^2}$,where $u = e^{\sin x} + \sinh^3 x$ and $v = \cosh x - 	an x$.First,find the derivatives at $x = 0$:$u(0) = e^{\sin 0} + \sinh^3 0 = e^0 + 0 = 1$.$u'(x) = e^{\sin x} \cos x + 3 \sinh^2 x \cosh x$.$u'(0) = e^0 \cos 0 + 3 \sinh^2 0 \cosh 0 = 1 \cdot 1 + 0 = 1$.$v(0) = \cosh 0 - 	an 0 = 1 - 0 = 1$.$v'(x) = \sinh x - \sec^2 x$.$v'(0) = \sinh 0 - \sec^2 0 = 0 - 1 = -1$.Now,substitute these values into the quotient rule formula at $x = 0$:$y'(0) = \frac{u'(0)v(0) - u(0)v'(0)}{(v(0))^2} = \frac{(1)(1) - (1)(-1)}{(1)^2} = \frac{1 + 1}{1} = 2$.Thus,the correct option is $D$.

If $y = \frac{e^{\sin x} + \sinh^3 x}{\cosh x - \tan x}$,then $y^{\prime}(0) = $

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