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

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(D) दिया गया है $f(x) = \frac{e^{2x} - e^{-2x}}{e^{3x} + e^{-3x}}$.भागफल नियम $\left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{

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$2$

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(D) दिया गया है $f(x) = \frac{e^{2x} - e^{-2x}}{e^{3x} + e^{-3x}}$.भागफल नियम $\left( \frac{u}{v} \right)^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{v^2}$ का उपयोग करते हुए:मान लीजिए $u = e^{2x} - e^{-2x}$ और $v = e^{3x} + e^{-3x}$.तब $u^{\prime} = 2e^{2x} + 2e^{-2x}$ और $v^{\prime} = 3e^{3x} - 3e^{-3x}$.$x = 0$ पर:$u(0) = e^0 - e^0 = 1 - 1 = 0$.$v(0) = e^0 + e^0 = 1 + 1 = 2$.$u^{\prime}(0) = 2(1) + 2(1) = 4$.$v^{\prime}(0) = 3(1) - 3(1) = 0$.अब, $f^{\prime}(0) = \frac{u^{\prime}(0)v(0) - u(0)v^{\prime}(0)}{(v(0))^2}$.$f^{\prime}(0) = \frac{(4)(2) - (0)(0)}{(2)^2} = \frac{8}{4} = 2$.

यदि $f(x) = \frac{e^{2x} - e^{-2x}}{e^{3x} + e^{-3x}}$ है, तो $f^{\prime}(0) = $

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