Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (i)/Solution 1

Using the fact that

${\frac {d}{dx}}\sin ^{-1}(x)={\frac {1}{\sqrt {1-x^{2}}}},$

and the chain rule, we get

${\begin{aligned}{\frac {d}{dx}}y&={\frac {d}{dx}}\sin ^{-1}(3x+1)\\&={\frac {1}{\sqrt {1-(3x+1)^{2}}}}\cdot {\frac {d}{dx}}(3x+1)\\&={\frac {3}{\sqrt {1-(3x+1)^{2}}}}.\end{aligned}}$