Derivative of square root of sin x using first principle

In this section, we will learn how to find

1. the derivative of the square root of sin x by definition or
2. 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:

$f^{\prime }(x)=\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}$

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

$f^{\prime }(x)=\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)-f(x)}{h}}$

$=\underset{h\to 0}{lim}{\displaystyle \frac{\sqrt{\sin (x+h)}-\sqrt{\sin x}}{h}}$

Rationalizing the numerator, $f^{\prime }(x)$ is equal to

$=\underset{h\to 0}{lim}[{\displaystyle \frac{\sqrt{\sin (x+h)}-\sqrt{\sin x}}{h}}\times$ ${\displaystyle \frac{\sqrt{\sin (x+h)}+\sqrt{\sin x}}{\sqrt{\sin (x+h)}+\sqrt{\sin x}}}]$

$=\underset{h\to 0}{lim}{\displaystyle \frac{\sin (x+h)-\sin x}{h\sqrt{\sin (x+h)}+\sqrt{\sin x}}}$

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

$=\underset{h\to 0}{lim}{\displaystyle \frac{\sin x\cos h+\cos x\sin h-\sin x}{h\sqrt{\sin (x+h)}+\sqrt{\sin x}}}$

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

$=\underset{h\to 0}{lim}{\displaystyle \frac{\sin x(\cos h-1)+\cos x\sin h}{h\sqrt{\sin (x+h)}+\sqrt{\sin x}}}$

$=\underset{h\to 0}{lim}{\displaystyle \frac{\sin x(\cos h-1)}{h\sqrt{\sin (x+h)}+\sqrt{\sin x}}}$ $+\underset{h\to 0}{lim}{\displaystyle \frac{\cos x\sin h}{h\sqrt{\sin (x+h)}+\sqrt{\sin x}}}$

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

$=\underset{h\to 0}{lim}{\displaystyle \frac{\cos x\sin h}{h\sqrt{\sin (x+h)}+\sqrt{\sin x}}}$

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

$=1\times {\displaystyle \frac{\cos x}{\sqrt{\sin x}+\sqrt{\sin x}}}$ $[\because \underset{h\to 0}{lim}{\displaystyle \frac{\sin h}{h}}=1]$

$={\displaystyle \frac{\cos x}{2\sqrt{\sin x}}}$ 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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