Question

compute the derivative of :-
f(x)=x+1/x​

Answer 1

Formula Used :-

$\boxed{ \red{ \bf \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

$\boxed{ \red{ \bf \: \dfrac{d}{dx} {x} = 1}}$

Solution :-

$\rm :\longmapsto\:Let \: y \: = \: x + \dfrac{1}{x}$

On differentiating w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} \: y \: = \dfrac{d}{dx} ( \: x + \dfrac{1}{x} )$

$\rm :\longmapsto\:\dfrac{dy}{dx} \: \: = \dfrac{d}{dx} \: x +\dfrac{d}{dx}( {x}^{ - 1})$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 1 + ( - 1) {x}^{ - 1 - 1}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 1 - {x}^{ - 2}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 1 - \dfrac{1}{ {x}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ {x}^{2} - 1}{ {x}^{2} }$

Additional Information :-

$\boxed{ \green{ \bf \: \dfrac{d}{dx}k = 0}}$

$\boxed{ \green{ \bf \: \dfrac{d}{dx}sinx = cosx}}$

$\boxed{ \green{ \bf \: \dfrac{d}{dx}cosx = - sinx}}$

$\boxed{ \green{ \bf \: \dfrac{d}{dx}cosecx = - cosecx \: cotx}}$

$\boxed{ \green{ \bf \: \dfrac{d}{dx}secx = secx \: tanx}}$

$\boxed{ \green{ \bf \: \dfrac{d}{dx} tanx = {sec}^{2}x}}$

$\boxed{ \green{ \bf \: \dfrac{d}{dx} cotx = - {cosec}^{2}x}}$

$\boxed{ \green{ \bf \: \dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x} }}}$

$\boxed{ \green{ \bf \: \dfrac{d}{dx} u.v = \: v\dfrac{d}{dx} u \: + \: u\dfrac{d}{dx} v}}$

$\boxed{ \green{ \bf \: \dfrac{d}{dx} \dfrac{u}{v} = \dfrac{v\dfrac{d}{dx} u \: - \: u\dfrac{d}{dx} v}{ {v}^{2} }}}$

Note :-

Let us solved one example using quotient rule :-

$\rm :\longmapsto\:Let \: y \: = \: \dfrac{x + 1}{x}$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} \: y \: = \dfrac{d}{dx} \: \dfrac{x + 1}{x}$

Here,

• u = x + 1

• v = x

On substituting these values of u and v, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{x\dfrac{d}{dx} (x + 1) - (x + 1)\dfrac{d}{dx} x}{ {x}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{x(1 + 0) - (x + 1)(1)}{ {x}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{x - x - 1}{ {x}^{2} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{-1}{ {x}^{2} }$