Question

If $f\left(x\right)=\left(lo{g}_{cotx}tanx\right)(lo{g}_{tanx}cotx{)}^{-1}+ta{n}^{-1}(\frac{x}{\sqrt{4-x^2}})$ , then f’(0) is equal to

• A. $-2$
• B. $2$
• C. $\frac{1}{2}$
• D. $0$

Answer 1

The correct option is C $\frac{1}{2}$

$f\left(x\right)=\left(lo{g}_{cotx}tanx\right)(lo{g}_{tan}xcotx{)}^{-1}+ta{n}^{-1}(\frac{x}{\sqrt{4-x^2}})=(-1\left)\right(-1{)}^{-1}+ta{n}^{-1}\left(\frac{x}{\sqrt{4-x^2}}\right)=1+ta{n}^{-1}\left(\frac{x}{\sqrt{4-x^2}}\right)\therefore f^{\prime }\left(0\right)=2^{-1}$