# Use the chain rule to find d/dx[tan(sqrtx)] ?

Jan 11, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\sec}^{2} \sqrt{x}}{2 \sqrt{x}}$

#### Explanation:

$y = \tan \sqrt{x}$

Apply the chain rule and standard differential.

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \sqrt{x} \cdot \frac{d}{\mathrm{dx}} \sqrt{x}$

Apply power rule.

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} \sqrt{x} \cdot \frac{1}{2} {x}^{- \frac{1}{2}}$

$= \frac{{\sec}^{2} \sqrt{x}}{2 \sqrt{x}}$

Jan 11, 2018

$\frac{d}{\mathrm{dx}} \left[\tan \left(\sqrt{x}\right)\right] = {\sec}^{2} \frac{\sqrt{x}}{2 \sqrt{x}}$

#### Explanation:

Apply the chain rule:

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

Let $f \left(x\right) = \tan \left(x\right)$ and $g \left(x\right) = \sqrt{x}$

Thus, $f ' \left(x\right) = {\sec}^{2} x$ and $g ' \left(x\right) = \frac{1}{2 \sqrt{x}}$

So

$\frac{d}{\mathrm{dx}} \left[\tan \left(\sqrt{x}\right)\right] = {\sec}^{2} \left(\sqrt{x}\right) \cdot \frac{1}{2 \sqrt{x}} = {\sec}^{2} \frac{\sqrt{x}}{2 \sqrt{x}}$