Let y = left(frac{3^{x}-1}{3^{x}+1}right) sin | vedclass

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(A) Given, $y = \left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x + \log_{e}(1+x)$.Applying the product rule to the first term and the chain rule to the sec

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(A) Given, $y = \left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x + \log_{e}(1+x)$.Applying the product rule to the first term and the chain rule to the second term:$\frac{dy}{dx} = \left(\frac{3^{x}-1}{3^{x}+1}\right) \frac{d}{dx}(\sin x) + \sin x \frac{d}{dx}\left(\frac{3^{x}-1}{3^{x}+1}\right) + \frac{1}{1+x}$.Using the quotient rule for $\frac{d}{dx}\left(\frac{3^{x}-1}{3^{x}+1}\right)$:$\frac{d}{dx}\left(\frac{3^{x}-1}{3^{x}+1}\right) = \frac{(3^{x}+1)(3^{x} \ln 3) - (3^{x}-1)(3^{x} \ln 3)}{(3^{x}+1)^{2}} = \frac{3^{x} \ln 3 (3^{x} + 1 - 3^{x} + 1)}{(3^{x}+1)^{2}} = \frac{2 \cdot 3^{x} \ln 3}{(3^{x}+1)^{2}}$.Substituting this back into the derivative expression:$\frac{dy}{dx} = \left(\frac{3^{x}-1}{3^{x}+1}\right) \cos x + \sin x \left(\frac{2 \cdot 3^{x} \ln 3}{(3^{x}+1)^{2}}\right) + \frac{1}{1+x}$.Evaluating at $x = 0$:$\left(\frac{dy}{dx}\right)_{x=0} = \left(\frac{3^{0}-1}{3^{0}+1}\right) \cos(0) + \sin(0) \left(\frac{2 \cdot 3^{0} \ln 3}{(3^{0}+1)^{2}}\right) + \frac{1}{1+0}$.$= \left(\frac{1-1}{1+1}\right) \cdot 1 + 0 \cdot \left(\frac{2 \ln 3}{4}\right) + 1 = 0 + 0 + 1 = 1$.

Let $y = \left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x + \log_{e}(1+x)$ for $x > -1$. Then, at $x = 0$, $\frac{dy}{dx}$ equals:

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