# What is the derivative of this function y=sec^-1(4x)?

Oct 6, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\left\mid x \right\mid \sqrt{16 {x}^{2} - 1}}$

#### Explanation:

You may already know that $\frac{d}{\mathrm{dx}} {\sec}^{-} 1 \left(x\right) = \frac{1}{\left\mid x \right\mid \sqrt{{x}^{2} - 1}}$. Then, according to the chain rule, we see that we will be following the rule:

$\frac{d}{\mathrm{dx}} {\sec}^{-} 1 \left(f \left(x\right)\right) = \frac{1}{\left\mid f \left(x\right) \right\mid \sqrt{{\left(f \left(x\right)\right)}^{2} - 1}} \cdot f ' \left(x\right)$

So, for ${\sec}^{-} 1 \left(4 x\right)$, we see that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\left\mid 4 x \right\mid \sqrt{{\left(4 x\right)}^{2} - 1}} \cdot \frac{d}{\mathrm{dx}} 4 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4}{\left\mid 4 x \right\mid \sqrt{16 {x}^{2} - 1}}$

Note that $\left\mid 4 x \right\mid = 4 \left\mid x \right\mid$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\left\mid x \right\mid \sqrt{16 {x}^{2} - 1}}$