Question

$\bold \red{Differentiate \: y= 5x^{-4}}$


❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
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Answer 1

Given : $y=5x^{-4}$

To find : Differentiation of y w.r.t. x

Solution :

$\implies y = 5 {x}^{ - 4}$

Differentiating both sides w.r.t. x, we get :

$\implies \dfrac{dy}{dx} = \dfrac{d}{dx} ( 5 {x}^{ - 4} )$

$\implies \dfrac{dy}{dx} =5 \times \dfrac{d}{dx} ( {x}^{ - 4} )$

We know that,

• $\boxed{ \dfrac{d}{dx} ( {x}^{n} ) = n{x}^{n - 1} }$

By applying this, we get :

${ \implies \dfrac{dy}{dx} =5 \times ( - 4) \times {x}^{ - 4 - 1} }$

${ \implies \dfrac{dy}{dx} = - 20 \times {x}^{ - 5} }$

$\underline{ \boxed{ \implies \dfrac{dy}{dx} = \dfrac{ - 20}{ {x}^{5} }}}$

Hence this is the required result.

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Answer 2

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

Given function is

$\rm :\longmapsto\:y = {5x}^{ - 4}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} {5x}^{ - 4}$

We know,

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 5\dfrac{d}{dx} {x}^{ - 4}$

We know,

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

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 5 \times ( - 4) {x}^{ - 4 - 1}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - 20{x}^{ - 5}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: \dfrac{20}{ {x}^{5} }$

Hence,

$\purple{\rm \implies\:\boxed{ \bf{ \: \dfrac{dy}{dx} = - \: \frac{20}{ {x}^{5} } \: }}}$

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More to know :-

$\green{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$