If f(x) = frac{e^{2x} - e^{-2x}}{e^{3x} + e^{ | vedclass

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(D) Given $f(x) = \frac{e^{2x} - e^{-2x}}{e^{3x} + e^{-3x}}$.Using the quotient rule $\left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime}v - uv^{\p

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(D) Given $f(x) = \frac{e^{2x} - e^{-2x}}{e^{3x} + e^{-3x}}$.Using the quotient rule $\left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$:Let $u = e^{2x} - e^{-2x}$ and $v = e^{3x} + e^{-3x}$.Then $u^{\prime} = 2e^{2x} + 2e^{-2x}$ and $v^{\prime} = 3e^{3x} - 3e^{-3x}$.At $x = 0$:$u(0) = e^0 - e^0 = 1 - 1 = 0$.$v(0) = e^0 + e^0 = 1 + 1 = 2$.$u^{\prime}(0) = 2(1) + 2(1) = 4$.$v^{\prime}(0) = 3(1) - 3(1) = 0$.Now, $f^{\prime}(0) = \frac{u^{\prime}(0)v(0) - u(0)v^{\prime}(0)}{(v(0))^2}$.$f^{\prime}(0) = \frac{(4)(2) - (0)(0)}{(2)^2} = \frac{8}{4} = 2$.

If $f(x) = \frac{e^{2x} - e^{-2x}}{e^{3x} + e^{-3x}}$, then $f^{\prime}(0) = $

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