Question

If $y={\sin }^{-1}\left(3^{-x}\right)$, then $\frac{dy}{dx}=$

• A. $\frac{-\log 3}{\sqrt{3^{2x}-1}}$
• B. $\frac{3^x\log 3}{\sqrt{3^{2x}-1}}$
• C. $\frac{-3^{-x}\log 3}{\sqrt{3^{2x}-1}}$
• D. $\frac{\log 3}{3^x\sqrt{3^{2x}-1}}$

Answer 1

The correct option is A $\frac{-\log 3}{\sqrt{3^{2x}-1}}$

$y={\sin }^{-1}\left(3^{-x}\right)$

$\frac{dy}{dx}=\frac{1}{\sqrt{1-{\left(3^{-x}\right)}^2}}\frac{d}{dx}\left(3^{-x}\right)$

$=\frac{1}{\sqrt{1-3^{-2x}}}(-3^{-x}\log 3)$

$=\frac{1}{\sqrt{1-\frac{1}{3^{2x}}}}(-3^{-x}\log 3)$

$=\frac{-3^{-x}\log 3}{\sqrt{\frac{3^{2x}-1}{3^{2x}}}}$

$=\frac{-3^{x-x}\log 3}{\sqrt{3^{2x}-1}}$

$=\frac{-\log 3}{\sqrt{3^{2x}-1}}$