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

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