Question

If y = log₁₀(sin^–1x²) then at x = $\frac{1}{\sqrt{2}},\frac{dy}{dx}$ =

• A. $\frac{2\sqrt{6}\log_{10}e}{\pi}$
• B. $\frac{4\sqrt{6}\log_{10}e}{\pi}$
• C. $\frac{2\sqrt{6}\log_{e}10}{\pi}$
• D. $\frac{4\sqrt{6}\log_{e}10}{\pi}$

Answer 1

Answer: (B)

$\frac{4\sqrt{6}\log_{10}e}{\pi}$

Answer 2

$\frac{dy}{dx}$ = $\frac { 1 } { \log _ { \mathrm { e } } 10 } \left[ \frac { 1 } { \sin ^ { - 1 } \mathrm { x } ^ { 2 } } \right] \cdot \frac { 1 } { \sqrt { 1 - \mathrm { x } ^ { 4 } } } \cdot 2 \mathrm { x }$ $\left\lbrack \frac{dy}{dx} \right\rbrack_{x = \frac{1}{\sqrt{2}}}$

= $\frac{1}{\log_{e}10}\left\lbrack \frac{1}{\sin^{- 1}\frac{1}{2}} \right\rbrack.\frac{1}{\sqrt{1 - \frac{1}{4}}}.2 \times \frac{1}{\sqrt{2}}$

= $\frac{\log_{10}e}{\pi} \times \frac{4 \times \sqrt{3}}{2} \times \sqrt{2}$ = $\frac{4\sqrt{6}}{\pi}\log_{10}e$