Browse Questions

# Find$\Large\frac{dy}{dx}\;\normalsize when\;y=\sin ^{-1} \bigg[\Large\frac{5x+12 \sqrt{1-x^2}}{13}\bigg]$

Toolbox:
• $sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2} )=sin^{-1}x+sin^{-1}y$
Take x as x and y as $\large\frac{12}{13}$ in the formula
$y=sin^{-1} \bigg[ x \sqrt{1-\large\frac{144}{169}}+\large\frac{12}{13}\sqrt{1-x^2} \bigg]$
$= sin^{-1}x+sin^{-1}\large\frac{12}{13}$

$\large\frac{dy}{dx}=\large\frac{1}{\sqrt{1-x^2}}$

edited Mar 26, 2013