Question

find the derivative of a/x^4 - b/x^2 + cos x​

Answer 1

Identities Used :-

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

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

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:Let \: y = \dfrac{a}{ {x}^{4} } - \dfrac{b}{ {x}^{2} } + cosx$

$\rm :\longmapsto\:y = {ax}^{ - 4} - {bx}^{ - 2} + cosx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}( {ax}^{ - 4} - {bx}^{ - 2} + cosx)$

$\rm :\longmapsto\:\dfrac{dy}{dx}= \dfrac{d}{dx}{ax}^{ - 4} - \dfrac{d}{dx}{bx}^{ - 2} + \dfrac{d}{dx}cosx$

$\rm :\longmapsto\:\dfrac{dy}{dx}= a\dfrac{d}{dx}{x}^{ - 4} - b\dfrac{d}{dx}{x}^{ - 2} - sinx$

$\rm :\longmapsto\:\dfrac{dy}{dx} = a( - 4 {x}^{( - 4 - 1)}) - b( - 2 {x}^{(- 2 - 1)}) - sinx$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - 4a {x}^{ - 5} + 2 b{x}^{ -3} - sinx$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \dfrac{4a}{ {x}^{5}} + \dfrac{2b}{ {x}^{3} } - sinx$

Additional Information :-

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

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

$\boxed{ \red{ \bf \: \dfrac{d}{dx} {a}^{x} = {a}^{x} log(a)}}$

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

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

$\boxed{ \red{ \bf \: \dfrac{d}{dx}k \: f(x) = k \: \dfrac{d}{dx}f(x)}}$

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

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