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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\frac{d}{d x} ((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)

y = (6 x^{2} - 6 x + 3) (1 + 2 x), at x = 1

\int e^{y} y - e^{y} d y

\int \frac{1}{e ^{x} e ^{\frac{1}{x ^{2}}}} d x

\int \frac{x ^{2} - x + 1}{x} d x

(x - 9) y^{^{'}} + y = x^{2} - 10

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)