Answer
${dy}={2{x}}\sec({{x}^2-1})\tan({{x}^2-1})){dx}$
Work Step by Step
We evaluate the function: $y=\sec({{x}^2-1})$
on differentiating the above:
$\frac{dy}{dx}=\frac{d\sec({{x}^2-1})}{dx}$
or $\frac{dy}{dx}=\sec({{x}^2-1})\tan({{x}^2-1})\frac{d(({x})^2-1)}{dx}$
or $\frac{dy}{dx}={2{x}}\sec({{x}^2-1})\tan({{x}^2-1}))$
or ${dy}={2{x}}\sec({{x}^2-1})\tan({{x}^2-1})){dx}$
The final answer is: ${dy}={2{x}}\sec({{x}^2-1})\tan({{x}^2-1})){dx}$
