Derivative of square root of sin x using first principle

In this section, we will learn how to find

the derivative of the square root of sin x by definition or
the derivative of the square root of sin x from first principle.

To answer the question, let us first know the definition of the derivative.

Definition of derivative: Let $f(x)$ be a differentiable function of $x$. From first principle of by definition, the derivative of $f(x)$ is given as follows:

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

Table of Contents

Derivative of the Square Root of Sin x from first principle:

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

Solution:

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

So by the definition or from the first principle, we get that

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

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

Rationalizing the numerator, $f'(x)$ is equal to

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

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

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

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

$[\because \sin(a+b)=\sin a \cos b+\cos a \sin b]$

$=\lim\limits_{h \to 0}\dfrac{\sin x (\cos h-1)+\cos x \sin h}{h \sqrt{\sin(x+h)}+\sqrt{\sin x}}$

$=\lim\limits_{h \to 0}\dfrac{\sin x (\cos h-1)}{h \sqrt{\sin(x+h)}+\sqrt{\sin x}}$ $+\lim\limits_{h \to 0}\dfrac{\cos x \sin h}{h \sqrt{\sin(x+h)}+\sqrt{\sin x}}$

$=0+\lim\limits_{h \to 0}\dfrac{\cos x \sin h}{h \sqrt{\sin(x+h)}+\sqrt{\sin x}}$ $[\because \lim\limits_{h \to 0} \dfrac{\cos h-1}{h}=0]$

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

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

$=1 \times \dfrac{\cos x}{\sqrt{\sin x}+\sqrt{\sin x}}$ $[\because \lim\limits_{h \to 0} \dfrac{\sin h}{h}=1]$

$=\dfrac{\cos x}{2\sqrt{\sin x}}$ ans.

So the derivative of square root of sinx is equal to (cos x)/(2 root sin x), obtained by the first principle of derivatives, that is, the limit definition of derivatives.

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