sanuluy

2021-01-19

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

lamusesamuset

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
$\frac{d}{dx}\left[{\left(2x\right)}^{4x}\right]=2\left[{e}^{4x\mathrm{ln}\left(x\right)}\right]\frac{d}{dx}\left[4x\mathrm{ln}\left(x\right)\right]$
$=2\left[{e}^{4x\mathrm{ln}\left(x\right)}\right]4\left[x×\frac{1}{x}+\mathrm{ln}\left(x\right)\right]$
$=2\left[{e}^{4x\mathrm{ln}\left(x\right)}\right]\left[1+\mathrm{ln}\left(x\right)\right]$
$2{x}^{4x}\left[1+l\left(x\right)\right]$

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