Stacie Worsley

2022-01-14

Derivatives Evaluate the following derivatives.
$\frac{d}{dx}\left({\left(2x\right)}^{4x}\right)$

Ethan Sanders

Expert

Step 1
Given,
$\frac{d}{dx}\left[{\left(2x\right)}^{4x}\right]$
Step 2
Given equation can be written as
$\frac{d}{dx}\left[{\left(2x\right)}^{4x}\right]=2\frac{d}{dx}\left[{\left(x\right)}^{4x}\right]$
We can write ${x}^{4x}$ as follows
${x}^{4x}={e}^{\mathrm{ln}\left({x}^{4x}\right)}$
$={e}^{4x\mathrm{ln}\left(x\right)}$
Hence
$\frac{d}{dx}\left[{\left(2x\right)}^{4x}\right]=2\frac{d}{dx}\left[{e}^{4x\mathrm{ln}\left(x\right)}\right]$
We apply the chain rule
$\frac{d}{dx}\left[{\left(2x\right)}^{4x}\right]=2\left[{e}^{4x}\mathrm{ln}\left(x\right)\right]\frac{d}{dx}\left[4x\mathrm{ln}\left(x\right)\right]$
$=2\left[{e}^{4x}\mathrm{ln}\left(x\right)\right]4\left[x×\frac{1}{x}+\mathrm{ln}\left(x\right)\right]$
$=2\left[{e}^{4x\mathrm{ln}\left(x\right)}\right]\left[1+\mathrm{ln}\left(x\right)\right]$
$=2{x}^{4x}\left[1+\mathrm{ln}\left(x\right)\right]$

Neunassauk8

Expert

${\left({\left(2x\right)}^{4x}\right)}^{\prime }={\left({\left({e}^{\mathrm{ln}2x}\right)}^{4x}\right)}^{\prime }$ [$2x={e}^{\mathrm{ln}2x}$]
$={\left({e}^{4x}\mathrm{ln}2x\right)}^{\prime }$ [Power rule]
$={e}^{4x\mathrm{ln}2x}\cdot {\left(4x\mathrm{ln}2x\right)}^{\prime }$ [Chain rule]
$={e}^{4x\mathrm{ln}2x}\cdot \left(4\mathrm{ln}2x+4x\cdot \frac{1}{2x}\cdot 2\right)$ [Product rule + chain rule]
$={\left(2x\right)}^{4x}\left(4\mathrm{ln}2x+4\right)$
Result:
${\left(2x\right)}^{4x}\left(4\mathrm{ln}2x+4\right)$

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