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.

Derivative of root cos x using first principle

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.


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


Q1: What is the derivative of root cosx?

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

Spread the love

Leave a Comment