Differentiate x x from first principle.

Question

Differentiate $\frac{\cos x}{x}$ from first principle.

✓

Answer

We have to find the derivative of $f(x)=\frac{\omega s x}{x}$
Derivative of a function $f ( x )$ is given by $f ^{\prime}( x )=\underset{{h \rightarrow 0}}{\lim}=\frac{f(x+h)-f(x)}{h}$ (where $h$ is a very small positive number)
$\therefore$ Derivative of $f ( x )=\frac{\cos x}{x}$ is given as $f ^{\prime}( x )=\underset{{h \rightarrow 0}}{\lim}=\frac{f(x+h)-f(x)}{h}$
$\Rightarrow f ( x )=\underset{{h \rightarrow 0}}{\lim} \frac{\frac{\frac{\cos (x+h)}{z+h}-\frac{\cos z}{z}}{h}}{h}$
$\Rightarrow f ( x )=\lim _{ h \rightarrow 0} \frac{\frac{z \cos (x+h)-(x+h) \cos x}{z(x+h)}}{h}=\underset{{h \rightarrow 0}}{\lim} \frac{x \cos (x+h)-(x+h) \cos x}{h(x)(x+h)}$
Using the algebra of limits we have
$\Rightarrow f ( x )=\lim _{ h \rightarrow 0} \frac{x \cos (x+h)-(x+h) \cos x}{h} \times \underset{{h \rightarrow 0}}{\lim} \frac{1}{x(x+h)}$
$\Rightarrow f ( x )=\lim _{ h \rightarrow 0} \frac{x \cos (x+h)-(x+h) \cos x}{h} \times \frac{1}{x(x+0)}$
$\Rightarrow f ( x )=\frac{1}{ x ^2} \lim _{ h \rightarrow 0} \frac{x \cos (x+h)-(x+h) \cos x}{h}$
$\Rightarrow f ( x )=\frac{1}{ x ^2} \lim _{ h \rightarrow 0} \frac{x \cos (x+h)-x \cos x-h \cos x}{h}$
Using the algebra of limits, we have:
$\Rightarrow f ( x )=\frac{1}{ x ^2}\left\{\underset{{h \rightarrow 0}}{\lim} \frac{-h \cos x}{h}+\underset{{h \rightarrow 0}}{\lim} \frac{x \cos (x+h)-x \cos x}{h}\right\}$
$\Rightarrow f ( x )=\frac{1}{ x ^2}\left\{-\underset{{h \rightarrow 0}}{\lim} \cos x+\underset{{h \rightarrow 0}}{\lim} \frac{x(\cos (x+h)-\cos x)}{h}\right\}$
Using the algebra of limits we have:
$\therefore f ^{\prime}( x )=-\frac{\cos x}{x^2}+\frac{1}{x} \underset{{h \rightarrow 0}}{\lim} \frac{\cos (x+h)-\cos x}{h}$
We can't evaluate the limits at this stage only as on putting value it will take $\frac{0}{0}$ form.
So, we need to do little modifications.
Use: $\cos A -\cos B =-2 \sin \left(\frac{(A+B)}{2}\right) \sin \left(\frac{(A-B)}{2}\right)$
$\therefore f^{\prime}(x)=-\frac{\cos x}{x^2}+\frac{1}{x} \underset{{h \rightarrow 0}}{\lim} \frac{-2 \sin \left(\frac{2 x+h}{2}\right) \sin \left(\frac{h}{2}\right)}{h}$
$\Rightarrow f^{\prime}(x)=-\frac{\cos x}{x^2}+\frac{1}{x} \underset{{h \rightarrow 0}}{\lim} \frac{\sin \left(x+\frac{h}{2}\right) \sin \left(\frac{h}{2}\right)}{\frac{h}{2}}$
Using algebra of limits:
$\Rightarrow f^{\prime}(x)=-\frac{\cos x}{x^2}+\frac{1}{x} \underset{{h \rightarrow 0}}{\lim} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} \times \underset{{h \rightarrow 0}}{\lim} \sin \left(x+\frac{h}{2}\right)$
By using the formula we get : $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
$\Rightarrow f^{\prime}(x)=-\frac{\cos x}{x^2}+\frac{1}{x} \underset{{h \rightarrow 0}}{\lim} \sin \left(x+\frac{h}{2}\right)$
Put the value of $h$ to evaluate the limit:
$\therefore f^{\prime}(x)=-\frac{\cos x}{x^2}+\frac{1}{x} \times \sin (x+0)=-\frac{\cos x}{x^2}-\frac{\sin x}{x}$
Hence,
Derivative of $f(x)=(\cos x) / x$ is $-\frac{\cos x}{x^2}-\frac{\sin x}{x}$