Question

find the derivative of the following :-

$\bigstar \boxed {\sf y = \dfrac{ {x}^{ \frac{7}{2} } + {x}^{ \frac{1}{3} } + x + 1}{ {x}^{ \frac{1}{2} } } } \bigstar$
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Answer 1

$\bigstar \large \purple{ \pmb{ \sf Question:- }}$

find the derivative of the following :-

$\bigstar \boxed {\sf y = \dfrac{ {x}^{ \frac{7}{2} } + {x}^{ \frac{1}{3} } + x + 1}{ {x}^{ \frac{1}{2} } } } \bigstar$

$\bigstar \large \purple{ \pmb{ \sf Required\:Answer:- }}$

❍ We can write the given equation as :-

$: \implies \sf y = \dfrac{x ^{ \frac{7}{2} } }{ {x}^{ \frac{1}{2} } } + \dfrac{ {x}^{ \frac{1}{3} } }{ {x}^{ \frac{1}{2} } } + \dfrac{ {x} }{ {x}^{ \frac{1}{2} } } + \dfrac{1}{ {x}^{ \frac{1}{2} } }$

$: \implies \sf y = {x}^{3} + {x}^{ \frac{ - 1}{6} } + {x}^{ \frac{1}{2} } + {x}^{ \frac{ - 1}{2} }$

Now, we are going to find the derivative of the above equation ...

❍ We already know that :-

For any equation like $\sf x^{n}$, where n is a real no.

$\leadsto \underline{ \boxed {\pink{ \sf \frac{dy}{dx} = n. {x}^{n - 1} }}} \bigstar$

❍ Let's calculate the derivative of this equation !

$: \implies \sf \dfrac{dy}{dx} = 3 {x}^{3 - 1} - \dfrac{ 1}{6} {x}^{ \frac{ - 1}{6} - 1 } + \dfrac{1}{2} {x}^{ \frac{1}{2} - 1 } - \frac{1}{6} {x}^{ \frac{ - 1}{2} - 1 }$

$: \implies \boxed {\pink{{\sf \dfrac{dy}{dx} = 3 {x}^{2} - \dfrac{1}{6} {x}^{ \frac{ - 7}{6} } + \dfrac{1}{2} {x}^{ \frac{ - 1}{2} } - \dfrac{1}{2} {x}^{ \frac{ - 3}{2} } }}} \bigstar$

And we are done :D

Answer 2

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

Given expression is

$\rm :\longmapsto\:y = \dfrac{ {\bigg(x\bigg) }^{\dfrac{7}{2}} + {\bigg(x\bigg) }^{\dfrac{1}{3} } + x + 1}{{\bigg(x\bigg) }^{\dfrac{1}{2} }}$

$\rm :\longmapsto\:y = \dfrac{{\bigg(x\bigg) }^{\dfrac{7}{2} }}{{\bigg(x\bigg) }^{\dfrac{1}{2} }} + \dfrac{{\bigg(x\bigg) }^{\dfrac{1}{3} }}{{\bigg(x\bigg) }^{\dfrac{1}{2} }} + \dfrac{x}{{\bigg(x\bigg) }^{\dfrac{1}{2} }} + \dfrac{1}{{\bigg(x\bigg) }^{\dfrac{1}{2} }}$

We know,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{ \sf{ \: {x}^{m} \div {x}^{n} = {x}^{m - n} }}}$

So using this, we get

$\rm :\longmapsto\:y = {\bigg(x\bigg) }^{\dfrac{7}{2} - \dfrac{1}{2}} + {\bigg(x\bigg) }^{\dfrac{1}{3} - \dfrac{1}{2}} + {\bigg(x\bigg) }^{1 - \dfrac{1}{2} } + {\bigg(x\bigg) }^{ - \dfrac{1}{2} }$

$\rm :\longmapsto\:y = {\bigg(x\bigg) }^{\dfrac{7 - 1}{2}} + {\bigg(x\bigg) }^{\dfrac{2 - 3}{6}} + {\bigg(x\bigg) }^{\dfrac{2 - 1}{2} } + {\bigg(x\bigg) }^{ - \dfrac{1}{2} }$

$\rm :\longmapsto\:y = {\bigg(x\bigg) }^{\dfrac{6}{2}} + {\bigg(x\bigg) }^{\dfrac{ - 1}{6}} + {\bigg(x\bigg) }^{\dfrac{1}{2} } + {\bigg(x\bigg) }^{ - \dfrac{1}{2} }$

$\rm :\longmapsto\:y = {\bigg(x\bigg) }^{3} + {\bigg(x\bigg) }^{\dfrac{ - 1}{6}} + {\bigg(x\bigg) }^{\dfrac{1}{2} } + {\bigg(x\bigg) }^{ - \dfrac{1}{2} }$

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

$\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx} {\bigg(x\bigg) }^{3} +\dfrac{d}{dx} {\bigg(x\bigg) }^{\dfrac{ - 1}{6}} + \dfrac{d}{dx}{\bigg(x\bigg) }^{\dfrac{1}{2} } + \dfrac{d}{dx}{\bigg(x\bigg) }^{ - \dfrac{1}{2} }$

We know,

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

So, using this we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = 3 {x}^{2} - \dfrac{1}{6}{\bigg(x\bigg) }^{ - \dfrac{1}{6} - 1} + \dfrac{1}{2}{\bigg(x\bigg) }^{\dfrac{1}{2} - 1 } - \dfrac{1}{2}{\bigg(x\bigg) }^{ - \dfrac{1}{2} - 1}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 3 {x}^{2} - \dfrac{1}{6}{\bigg(x\bigg) }^{ - \dfrac{7}{6}} + \dfrac{1}{2}{\bigg(x\bigg) }^{ - \dfrac{1}{2} } - \dfrac{1}{2}{\bigg(x\bigg) }^{ - \dfrac{3}{2}}$

Additional Information :-

$\red{ \boxed{ \sf{ \dfrac{d}{dx}k\: = 0}}}$

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

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

$\red{ \boxed{ \sf{ \dfrac{d}{dx} {a}^{x} \: = {a}^{x} loga \: \: where \: a > 0}}}$

$\red{ \boxed{ \sf{ \dfrac{d}{dx}logx\: = \frac{1}{x} }}}$

$\red{ \boxed{ \sf{ \dfrac{d}{dx} log_{a}(x) \: = \frac{1}{x \: loga} \: \: \: where \: a > 0}}}$

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

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

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

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