"Derivatives Evaluate the following derivatives. ddx((2x)4x)" solved and explained

Stacie Worsley

Answered question

2022-01-14

Derivatives Evaluate the following derivatives.

\frac{d}{d x} ({(2 x)}^{4 x})

Answer & Explanation

Ethan Sanders

Beginner2022-01-15Added 35 answers

Step 1
Given,

\frac{d}{d x} [{(2 x)}^{4 x}]

Step 2
Given equation can be written as

\frac{d}{d x} [{(2 x)}^{4 x}] = 2 \frac{d}{d x} [{(x)}^{4 x}]

We can write

x^{4 x}

as follows

x^{4 x} = e^{\ln (x^{4 x})}

= e^{4 x \ln (x)}

Hence

\frac{d}{d x} [{(2 x)}^{4 x}] = 2 \frac{d}{d x} [e^{4 x \ln (x)}]

We apply the chain rule

\frac{d}{d x} [{(2 x)}^{4 x}] = 2 [e^{4 x} \ln (x)] \frac{d}{d x} [4 x \ln (x)]

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

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

= 2 x^{4 x} [1 + \ln (x)]

Neunassauk8

Beginner2022-01-16Added 30 answers

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

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