Q106 / 160 · AP EAMCET 2019 (21 Apr Shift 1) · Mathematics

If $f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}\right)$, then $f^{\prime}(1)=$

  • A
    -log2
  • B
    log2
  • C
    1
  • Correct answer
    D
    -1

Solution

Given, $$ \begin{gathered} f(x)=\cot ^{-1}\left(\frac{x^x-x^{-x}}{2}\right) \\ y=\cot ^{-1}\left(\frac{x^{2 x}-1}{2 \cdot x^x}\right) \end{gathered} $$ Put $x^x=\tan \theta$ $$ \begin{aligned} \therefore \quad y & =\cot ^{-1}\left(\frac{\tan ^2 \theta-1}{2 \tan \theta}\right) \\ & =\cot ^{-1}(-\cot 2 \theta) \\ y & =\pi-\cot ^{-1}(\cot 2 \theta) \\ y & =\pi-2 \theta \\ y & =\pi-2 \tan ^{-1}\left(x^x\right) \\ \therefore \quad \frac{d y}{d x} & =-\frac{2}{1+x^{2 x}} \cdot x^x(1+\log x) \end{aligned} $$ So, $\left.\quad \frac{d y}{d x}\right|_{x=1}=\frac{-2}{1+(1)^2} \cdot 1^1(1+\log 1)$ $$ =-\frac{2}{2} \cdot 1=-1 $$

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

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