Question

Differentiate with respect to x ​

Answer 1

$\large \bold \red{ \frac{d}{dx} \bigg( \bigg( \frac{32 {}^{ \frac{1}{9}} 243 {}^{ \frac{8}{9} } }{243} \bigg) {}^{ \sqrt{x - 5} } \bigg)}$

Rewrite

Equivalent expressions

${\large \bold \red{ \frac{}{} \bigg( \big \frac{32 {}^{ \frac{1}{9}} 243 {}^{ \frac{8}{9} } }{243} \bigg) {}^{ \sqrt{x - 5} } = e {}^{ \sqrt{x - 5}In \bigg( \frac{32 {}^{\frac{1}{9} } 243 {}^{ \frac{8}{9} } }{243} \bigg) } }}$

This gives:

${ \boxed{ \large \bold \red{\frac{d}{dx}e {}^{ \sqrt{x - 5}In \bigg( \frac{32 {}^{\frac{1}{9} }243 {}^{ \frac{8}{9} } }{243} \bigg)}}}}$

Apply the chain rule to the term $\bold{e {}^{ \sqrt{x - 5}In ( \frac{32 {}^{\frac{1}{9} }243 {}^{ \frac{8}{9} } }{243})}}$

Recall the definition of chain rule

$\bold \red{\frac{d}{dx} f(g(x)) = f'(g(x)) \frac{d}{dx} g(x)}$

Outside function

$\bold \red{f(v) = {e}^{v}}$

Inside function

$\bold \red{g(x) = \sqrt{x - 5} }$

$\bold \red{g(x) = \sqrt{x - 5} In \bigg( \frac{32 {}^{ \frac{1}{9}}243 {}^{ \frac{8}{9} } }{243} \bigg)}$

Derivative of outside function

$\bold \red{\frac{d}{dv} f(v) = {e}^{v}}$

Apply composition

$\bold \red{f'(g(x)) = e {}^{ \sqrt{x - 5} In( \frac{32 {}^{ \frac{1}{9} } 243 {}^{ \frac{8}{9} } }{243} )}}$

Derivative of inside function

$\bold \red{\frac{d}{dx} g(x) = \frac{In \bigg( \frac{32 {}^{ \frac{1}{9} }243 {}^{ \frac{8}{9} } }{243} \bigg) }{2 \sqrt{x - 2} } }$

Put it all together

$\bold\red{\frac{d}{dx} f(g(x)) \frac{d}{dx} g(x) = e {}^{ \sqrt{x - 5} }}$

${\bold \red{\frac{d}{dx} f(g(x)) \frac{d}{dx} g(x) = e {}^{ \sqrt{x - 5} In( \frac{32 {}^{ \frac{1}{9} } 243 {}^{ \frac{8}{9} } }{243} )} \times \bigg(\frac{In \bigg( \frac{32 {}^{ \frac{1}{9} }243 {}^{ \frac{8}{9} } }{243} \bigg) }{2 \sqrt{x - 2} } \bigg)}}$

This gives:

${\boxed{\bold \red{\frac{e {}^{ \sqrt{x - 5}In( \frac{32 {}^{ \frac{1}{9} }243 {}^{ \frac{8}{9}}}{243} ) }In \bigg( \frac{32 {}^{ \frac{1}{9} }243 {}^{ \frac{8}{9} } }{243} \bigg)}{2 \sqrt{x - 5} }}}}$

Answer 2

Given Question :-

Evaluate the following

$\rm :\longmapsto\:\dfrac{d}{dx} {\bigg[\dfrac{ {32}^{ \frac{1}{9}} \times {243}^{ \frac{8}{9} } }{243} \bigg]}^{ \sqrt{x - 5} }$

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

Let assume that

$\rm :\longmapsto\:y = {\bigg[\dfrac{ {32}^{ \frac{1}{9}} \times {243}^{ \frac{8}{9} } }{243} \bigg]}^{ \sqrt{x - 5} }$

We know,

$\boxed{ \tt{ \: \frac{ {x}^{m} }{ {x}^{n} } = {x}^{m - n} \: }}$

So, using this identity we get

$\rm :\longmapsto\:y = {\bigg[\dfrac{ {32}^{ \frac{1}{9}} }{ {(243)}^{1 - \frac{8}{9} } } \bigg]}^{ \sqrt{x - 5} }$

$\rm :\longmapsto\:y = {\bigg[\dfrac{ {32}^{ \frac{1}{9}} }{ {(243)}^{\frac{9 - 8}{9} } } \bigg]}^{ \sqrt{x - 5} }$

$\rm :\longmapsto\:y = {\bigg[\dfrac{ {32}^{ \frac{1}{9}} }{ {(243)}^{\frac{1}{9} } } \bigg]}^{ \sqrt{x - 5} }$

We know

$\boxed{ \tt{ \: \frac{ {x}^{m} }{ {y}^{m} } = \bigg[\dfrac{x}{y} \bigg]^{m} \: }}$

So, using this, we get

$\rm :\longmapsto\:y = {\bigg[\dfrac{32}{243} \bigg]}^{\dfrac{ \sqrt{x - 5} }{9}}$

$\rm :\longmapsto\:y = {\bigg[\dfrac{ {2}^{5} }{ {3}^{5} } \bigg]}^{\dfrac{ \sqrt{x - 5} }{9}}$

$\rm :\longmapsto\:y = {\bigg[\dfrac{2}{3} \bigg]}^{\dfrac{5 \sqrt{x - 5} }{9}}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} {\bigg[\dfrac{2}{3} \bigg]}^{\dfrac{5 \sqrt{x - 5} }{9}}$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} {a}^{x} \: = \: {a}^{x} \: loga \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = {\bigg[\dfrac{2}{3} \bigg]}^{\dfrac{5 \sqrt{x - 5} }{9}} log\bigg[\dfrac{2}{3} \bigg] \: \dfrac{d}{dx}\bigg[\dfrac{5 \sqrt{x - 5} }{9} \bigg]$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{5}{9} \times {\bigg[\dfrac{2}{3} \bigg]}^{\dfrac{5 \sqrt{x - 5} }{9}} log\bigg[\dfrac{2}{3} \bigg] \: \dfrac{d}{dx} \sqrt{x - 5}$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{5}{9} \times {\bigg[\dfrac{2}{3} \bigg]}^{\dfrac{5 \sqrt{x - 5} }{9}} log\bigg[\dfrac{2}{3} \bigg] \: \dfrac{1}{2 \sqrt{x - 5} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{5}{18} \times {\bigg[\dfrac{2}{3} \bigg]}^{\dfrac{5 \sqrt{x - 5} }{9}} log\bigg[\dfrac{2}{3} \bigg] \: \dfrac{1}{\sqrt{x - 5} }$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{5}{18 \sqrt{x - 5} } \times {\bigg[\dfrac{2}{3} \bigg]}^{\dfrac{5 \sqrt{x - 5} }{9}} log\bigg[\dfrac{2}{3} \bigg]$

Therefore,

$\red{\boxed{ \tt{ \: \dfrac{dy}{dx} =\dfrac{5}{18 \sqrt{x - 5} } \times {\bigg[\dfrac{2}{3} \bigg]}^{\dfrac{5 \sqrt{x - 5} }{9}} log\bigg[\dfrac{2}{3} \bigg] \: }}}$

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More to know :-

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