Science:Math Exam Resources/Courses/MATH102/December 2013/Question B 01/Solution 1

Differentiate both sides of $\tan y=x$ with respect to $x$. (Don't forget to apply the chain rule!)

$\begin{array}{rl}\frac{d}{dx}\tan y& =\frac{d}{dx}x\\ {\sec }^{2}y\cdot \frac{dy}{dx}& =1\\ \frac{dy}{dx}& =\frac{1}{{\sec }^{2}y}.\end{array}$

Now apply the trig identity ${\sec }^{2}y=1+{\tan }^{2}y$ to get

$\begin{array}{r}\frac{dy}{dx}=\frac{1}{1+{\tan }^{2}y}=\frac{1}{1+x^2}\end{array}$

after noting that $\tan y=x⟹{\tan }^{}y=x^2$.