sanuluy

2021-01-19

Evaluate the following derivatives.

$\frac{d}{dx}\left[{\left(2x\right)}^{4x}\right]$

lamusesamuset

Skilled2021-01-20Added 93 answers

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

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

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

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

$2{x}^{4x}[1+l\left(x\right)]$

Given,

Step 2

Given equation can be written as

We can write

Hence

We apply the chain rule

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