The derivative of root secx is equal to (secx tanx)/(2√secx). In this post, we will learn how to find the derivative of root secx by first principles.

The formula of the derivative of square root of secx is given as follows:

$\dfrac{d}{dx}(\sqrt{\sec x})=\dfrac{\sec x\tan x}{2\sqrt{\sec x}}$.

## Derivative of square root of secx from first principle

**Question:** Find the derivative of $\sqrt{\sec x}$ from first principle.

**Solution:**

We will follow the below steps in order to find the derivative of square root of secx using first principle. The steps are as follows:

**Step 1:**

Let f(x) = $\sqrt{\sec x}$

The derivative of f(x) by first principle is given by the limit formula

$\dfrac{d}{dx}(f(x))$ = lim_{h→0} $\dfrac{f(x+h)-f(x)}{h}$.

**Step 2:**

Put f(x)=$\sqrt{\sec x}$. Therefore, the differentiation of root sec x from first principle will be given by

$\dfrac{d}{dx}(\sqrt{\sec x})$ = lim_{h→0} $\dfrac{\sqrt{\sec (x+h)}-\sqrt{\sec x}}{h}$

Let us multiply both the numerator and the denominator by √sec(x + h) + √secx. As a result, the above

= lim_{h→0} $\Big[ \dfrac{\sqrt{\sec (x+h)}-\sqrt{\sec x}}{h}$ $\times \dfrac{\sqrt{\sec (x+h)}+\sqrt{\sec x}}{\sqrt{\sec (x+h)}+\sqrt{\sec x}} \Big]$

= lim_{h→0} $\dfrac{(\sqrt{\sec (x+h)})^2-(\sqrt{\sec x})^2}{h(\sqrt{\sec (x+h)}+\sqrt{\sec x})}$, here we use the formula (a-b)(a+b)=a^{2}-b^{2}.

= lim_{h→0} $\Big[ \dfrac{\sec (x+h)-\sec x}{h}$ $\times \dfrac{1}{\sqrt{\sec (x+h)}+\sqrt{\sec x}} \Big]$ **…(∗)**

**Step 3:**

Now, we know that the limit lim_{h→0} $\dfrac{\sec (x+h)-\sec x}{h}$ $=\dfrac{d}{dx}(\sec x)$ = secx tanx, by the first principle of derivatives.

Applying the product rule of limits, it follows from **(∗)** that

$\dfrac{d}{dx}(\sqrt{\sec x})$ = lim_{h→0} $\dfrac{\sec (x+h)-\sec x}{h}$ × lim_{h→0} $\dfrac{1}{\sqrt{\sec (x+h)}+\sqrt{\sec x}}$

= secx tanx $\times \dfrac{1}{\sqrt{\sec (x+0)}+\sqrt{\sec x}}$

= secx tanx $\times \dfrac{1}{2\sqrt{\sec x}}$

= $\dfrac{\sec x \tan x}{2\sqrt{\sec x}}$

Thus, the derivative of square root of secx is equal to (secx tanx)/(2√secx), and this is proved from the first principle of derivatives.

**Related Topics:**

Derivative of $\sqrt{\sin x}$ from first principle

Derivative of $\sqrt{\cos x}$ from first principle

Derivative of $\sqrt{\tan x}$ from first principle

Derivative of $\sqrt{\sin x}$ by chain rule

Derivative of $\sqrt{\cos x}$ by chain rule

## FAQs

**Q1: What is the derivative of root secx?**

Answer: The derivative of square root of secx is equal to (secx tanx)/(2√secx).