Question

Differentiate $\frac{sinx}{x}$ using the first principle.

Answer 1

$Lety=\frac{sinx}{x}$

Let $\delta y$ be an increment in y, corresponding to an increment $\delta xinx.$

$Then,y+\delta y=\frac{sin(x+\delta x)}{(x+\delta x)}$

$\Rightarrow \delta y=\{\frac{sin(x+\delta x)}{(x+\delta x)}-\frac{sinx}{x}\}=\frac{\left\{xsin\right(x+\delta x)-(x+\delta x\left)sinx\right\}}{x(x+\delta x)}$

$\Rightarrow \frac{\delta y}{\delta x}=\frac{\left\{xsin\right(x+\delta x)-(x+\delta x\left)sinx\right\}}{x(x+\delta x).\delta x}$

$\Rightarrow \frac{dy}{dx}=\underset{\delta x\to 0}{lim}\frac{\delta y}{\delta x}$

$=\underset{\delta x\to 0}{lim}\frac{\left\{xsin\right(x+\delta x)-(x+\delta x\left)sinx\right\}}{x(x+\delta x).\delta x}$

$\underset{\delta x\to 0}{lim}\frac{x\left[sin\right(x+\delta x)-sinx]-\left(\delta x\right)sinx}{x(x+\delta x).\delta x}$ $=\underset{\delta x\to 0}{lim}\frac{x[2cos(x+\frac{\delta x}{2})sin\left(\frac{\delta x}{2}\right)-(\delta x\left)sinx\right]}{x(x+\delta x).\delta x}$

$=\{\underset{\delta x\to 0}{lim}cos(x+\frac{\delta x}{2}).\underset{\delta x\to 0}{lim}\frac{sin\left(\delta x/2\right)}{\left(\delta /2\right)}\underset{\delta x\to 0}{lim}\frac{1}{(x+\delta x)}-\underset{\delta x\to 0}{lim}\frac{sinx}{x(x+\delta x)}\}$

$=(cosx\times 1\times \frac{1}{x})-\frac{sinx}{x^2}=(\frac{cosx}{x}-\frac{sinx}{x^2})=\frac{(xcosx-sinx)}{x^2}$

$Hence,\frac{d}{dx}\left(\frac{sinx}{x}\right)=\frac{(xcosx-sinx)}{x^2}$