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

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

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

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

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

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

Solution steps

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

Simplify

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

= \frac{d}{d x} (1 6^{x} x^{4 x})

Apply the Product Rule:

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

f = 1 6^{x}, g = x^{4 x}

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

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

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

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

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

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

f (x) = x^{2} - 4 x + 3, (1, 0)

\int_{0}^{π} sin (x^{2}) d x

\int cot^{3} (x) d x

f (x) = x + 8, (8, 4)

\int \frac{2 x}{9 - 4 x ^{2}} d x

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