# Derivative of root cos x using first principle

The derivative of square root of cosx is equal to (-sin x)/(2 root(cos x)).In this post, we will learn how to find the derivative of root cosx by first principle, that is, by the limit definition.

The first principle of derivative is defined as follows:

Let $f(x)$ be a differentiable function of $x$. The derivative of $f(x)$ using the first principle is denoted by $f'(x)$ and it is given below.

$f'(x)=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ $\cdots (i)$

## Derivative of the Square Root of cos x from First Principle:

Question: Find the Derivative of  $\sqrt{\cos x}$ from first principle.

Solution:

Step 1: Let $f(x)=\sqrt{\cos x}$

Applying the above definition (i) of the first principle of derivatives, we get that

$f'(x)=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim\limits_{h \to 0} \dfrac{\sqrt{\cos(x+h)}-\sqrt{\cos x}}{h}$ $\cdots (ii)$

Step 2: We now rationalize the numerator of (ii). Thus $f'(x)$ is equal to

$=\lim\limits_{h \to 0}[\dfrac{\sqrt{\cos(x+h)}-\sqrt{\cos x}}{h} \times$ $\dfrac{\sqrt{\cos(x+h)}+\sqrt{\cos x}}{\sqrt{\cos(x+h)}+\sqrt{\cos x}}]$

$=\lim\limits_{h \to 0}\dfrac{\cos(x+h)-\cos x}{h \sqrt{\cos(x+h)}+\sqrt{\cos x}}$

$[\because (a-b)(a+b)=a^2-b^2]$

$=\lim\limits_{h \to 0}$ $\dfrac{-2 \sin(\frac{x+h+x}{2}) \sin(\frac{x+h-x}{2})}{h \sqrt{\cos(x+h)}+\sqrt{\cos x}}$

[ applying the formula $\cos x – \cos y =$ $-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}$]

$=\lim\limits_{h \to 0}$ $\dfrac{-2 \sin(\frac{2x+h}{2}) \sin(\frac{h}{2})}{h \sqrt{\cos(x+h)}+\sqrt{\cos x}}$

$=\lim\limits_{h \to 0}$ $[ \dfrac{-\sin(\frac{2x+h}{2})}{\sqrt{\cos(x+h)}+\sqrt{\cos x}} \times$ $\dfrac{\sin \frac{h}{2}}{\frac{h}{2}}]$

$=\lim\limits_{h \to 0} [\dfrac{-\sin(\frac{2x+h}{2})}{\sqrt{\cos(x+h)}+\sqrt{\cos x}} \times$ $\lim\limits_{h \to 0} \dfrac{\sin \frac{h}{2}}{\frac{h}{2}}]$

Let $t=\frac{h}{2}$. Then $t \to 0$ as $h \to 0$.

$=\lim\limits_{h \to 0} \dfrac{-\sin(\frac{2x+h}{2})}{\sqrt{\cos(x+h)}+\sqrt{\cos x}} \times$ $\lim\limits_{t \to 0} \dfrac{\sin t}{t}$

$=\dfrac{-\sin(\frac{2x+0}{2})}{\sqrt{\cos(x+0)}+\sqrt{\cos x}}$ $\times 1, \quad$ as the limit of (sin x)/x is 1 as x tends to 0.

$=\dfrac{-\sin x}{2\sqrt{\cos x}}$

So the derivative of root cos x is (-sin x)/(2 root(cos x)) ans.

Derivative of root sinx by First Principle

Derivative of log 3x

Derivative of 1/logx

## FAQs

Q1: What is the derivative of root cosx?

Answer: The derivative of root cosx is equal to (-sin x)/(2 root(cos x)).