Q105 / 160 · TS EAMCET 2023 (13 May Shift 1) · Mathematics

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

  • A
    $-1$
  • B
    $0$
  • C
    $1$
  • Correct answer
    D
    $2$

Solution

$\begin{aligned} f(x) & =\frac{e^{2 x}-e^{-2 x}}{e^{3 x}+e^{-3 x}} \\ \Rightarrow f^{\prime}(x) & =\frac{\left(e^{3 x}+e^{-3 x}\right)\left(2 e^{2 x}+2 e^{-2 x}\right)}{\left(e^{2 x}-e^{-2 x}\right)\left(3 e^{3 x}-3 e^{-3 x}\right)} \\ \Rightarrow f^{\prime}(0) & =\frac{(1+1)(2 \cdot 1+2 \cdot 1)-(1-1)(3 \cdot 1-3 \cdot 1)}{(1+1)^2} \\ & =\frac{8}{4}=2 .\end{aligned}$

Hence, the correct answer is option: 'D'.

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