Question

If $f\left(x\right)=\left(lo{g}_{cotx}tanx\right){\left(lo{g}_{tanx}cotx\right)}^{-1}$ $+ta{n}^{-1}\frac{4x}{4-x^2}$, then $f^{\prime }(2)$ is equal to

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

Answer: (A)

$f\left(x\right)=\left(lo{g}_{cotx}tanx\right){\left(lo{g}_{tanx}cotx\right)}^{-1}+ta{n}^{-1}\frac{4x}{4-x^2}$
$=\frac{logtanx}{logcotx}\cdot \frac{logtanx}{logcotx}+ta{n}^{-1}\left(\frac{4x}{4-x^2}\right)$
$=\frac{\left(logtan{x}^2\right)}{\left(-logtan{x}^2\right)}+ta{n}^{-1}\left(\frac{4x}{4-x^2}\right)$
$=1+ta{n}^{-1}\left(\frac{4x}{4-x^2}\right)$
$\therefore f^{\prime }\left(x\right)=\frac{1}{1+{\left(\frac{4x}{4-x^2}\right)}^2}\cdot \frac{\left(4-x^2\right)4-4x\left(-2x\right)}{{\left(4-x^2\right)}^2}$
$=\frac{16-4x^2+8x^2}{{\left(4-x^2\right)}^2+16x^2}$
$=\frac{4\left(4+x^2\right)}{{\left(4-x^2\right)}^2+{\left(4x\right)}^2}$
Hence, $f^{\prime }\left(2\right)=\frac{4\left(4+4\right)}{0+{\left(8\right)}^2}$
$=\frac{30}{64}=\frac{1}{2}$