Question

7 ^ { x + 1 / x } find dy / dx? explain step by step please​

Answer 1

Answer:

$\mathrm{ 7^{\big(x+\frac{1}{x}\big)}.log7.\bigg(1 - \dfrac{1}{x^2}\bigg)}$

Step-by-step explanation:

Given :

$\mathrm{y = 7^{\big(x+\frac{1}{x}\big)}}$

To find :

$\mathrm{\dfrac{dy}{dx}}$

Solution :

$\longmapsto \mathrm{y = 7^{\big(x+\frac{1}{x}\big)}}$

Applying log on both sides,

We know that,

$\boxed{\mathrm{log\; a^m = m\: loga}}$

$\implies \mathrm{logy = {\big(x+\frac{1}{x}\big)}log7}$

Differentiating with respect to x both sides,

We know that,

$\boxed{\begin{array}{cc} \mathrm{\dfrac{d}{dx}(logx) = \dfrac{1}{x} }\\\\\mathrm{\dfrac{d}{dx}(x^n) = nx^{n-1}}\end{array}}$

So, here,

$\implies \mathrm{\dfrac{1}{y}\dfrac{dy}{dx} = {\bigg(1-\dfrac{1}{x^2}\bigg)}log7}$

$\implies \mathrm{\dfrac{dy}{dx} = ylog7\bigg(1 - \dfrac{1}{x^2}\bigg)}$

$\therefore \underline{\boxed{\mathbf{\dfrac{dy}{dx} = 7^{\big(x+\frac{1}{x}\big)}.log7.\bigg(1 - \dfrac{1}{x^2}\bigg)}}}$

❖ Extra information

⟡ Some basic differentiations :

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

Hope it helps!!

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Answer 2

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

Given function is

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

Let assume that

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

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

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

We know that

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{d}{dx} {a}^{x} = {a}^{x}loga \: \: \: \: \forall \: a \: > \: 0}}}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{dy}{dx}= {\bigg( 7\bigg) }^{x + \dfrac{1}{x} } log7 \: \dfrac{d}{dx} \: \bigg[x + \dfrac{1}{x} \bigg]$

We know that

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{d}{dx}(u + v) = \dfrac{d}{dx}u + \dfrac{d}{dx}v \: }}}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{dy}{dx}= {\bigg( 7\bigg) }^{x + \dfrac{1}{x} } log7 \: \: \bigg[\dfrac{d}{dx}x + \dfrac{d}{dx}\dfrac{1}{x} \bigg]$

We know that

$\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }} \\ \\ \bf \: and \\ \\ \boxed{ \tt{ \: \dfrac{d}{dx} \frac{1}{ {x}^{n} } = \frac{ - n}{ {x}^{n + 1} } \: }}$

So, using these results, we get

$\rm :\longmapsto\:\dfrac{dy}{dx}= {\bigg( 7\bigg) }^{x + \dfrac{1}{x} } log7 \: \: \bigg[1 + \dfrac{ - 1}{ {x}^{1 + 1} } \bigg]$

$\rm :\longmapsto\:\dfrac{dy}{dx}= {\bigg( 7\bigg) }^{x + \dfrac{1}{x} } log7 \: \: \bigg[1 - \dfrac{1}{ {x}^{2} } \bigg]$

$\\ \bf :\longmapsto\:\dfrac{dy}{dx}= {\bigg( 7\bigg) }^{x + \dfrac{1}{x} } log7 \: \: \bigg[ \dfrac{ {x}^{2} - 1}{ {x}^{2} } \bigg]$

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More to Know :-

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