What is the derivative of (4x)^{2x} ?

The derivative of (4x)^{2x} is ln(2)*2^{4x+2}x^{2x}+2^{1+4x}x^{2x}(ln(x)+1)

What is the first derivative of (4x)^{2x} ?

The first derivative of (4x)^{2x} is ln(2)*2^{4x+2}x^{2x}+2^{1+4x}x^{2x}(ln(x)+1)

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derivative of

(4 x)^{2 x}

ln (2) \cdot 2^{4 x + 2} x^{2 x} + 2^{1 + 4 x} x^{2 x} (ln (x) + 1)

Solution steps

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

Simplify

(4 x)^{2 x} : 1 6^{x} x^{2 x}

= \frac{d}{d x} (1 6^{x} x^{2 x})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 1 6^{x}, g = x^{2 x}

= \frac{d}{d x} (1 6^{x}) x^{2 x} + \frac{d}{d x} (x^{2 x}) 1 6^{x}

\frac{d}{d x} (1 6^{x}) = ln (2) \cdot 2^{4 x + 2}

\frac{d}{d x} (x^{2 x}) = 2 x^{2 x} (ln (x) + 1)

= ln (2) \cdot 2^{4 x + 2} x^{2 x} + 2 x^{2 x} (ln (x) + 1) \cdot 1 6^{x}

2 x^{2 x} (ln (x) + 1) \cdot 1 6^{x} = 2^{1 + 4 x} x^{2 x} (ln (x) + 1)

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

n = 0 \sum \infty \frac{1}{n ^{1.1}}

\frac{d}{d x} (\frac{5}{x ^{3}} - \frac{2}{x ^{2}} - \frac{1}{x} + 200)

y = - 2 x^{- 4} + \frac{4}{x} - \frac{2 x ^{2}}{3}

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

\int x 3^{x} d x

What is the derivative of (4x)^{2x} ?
The derivative of (4x)^{2x} is ln(2)*2^{4x+2}x^{2x}+2^{1+4x}x^{2x}(ln(x)+1)
What is the first derivative of (4x)^{2x} ?
The first derivative of (4x)^{2x} is ln(2)*2^{4x+2}x^{2x}+2^{1+4x}x^{2x}(ln(x)+1)