Derivative of root secx by First Principle

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)=a²-b².

= 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:

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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

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