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

The derivative of (4x^{4x}) is ln(2)*2^{8x+3}x^{4x}+2^{2+8x}x^{4x}(ln(x)+1)

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

The first derivative of (4x^{4x}) is ln(2)*2^{8x+3}x^{4x}+2^{2+8x}x^{4x}(ln(x)+1)

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

ln (2) \cdot 2^{8 x + 3} x^{4 x} + 2^{2 + 8 x} x^{4 x} (ln (x) + 1)

Solution steps

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

Simplify

(4 x)^{4 x} : 25 6^{x} x^{4 x}

= \frac{d}{d x} (25 6^{x} x^{4 x})

Apply the Product Rule:

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

f = 25 6^{x}, g = x^{4 x}

= \frac{d}{d x} (25 6^{x}) x^{4 x} + \frac{d}{d x} (x^{4 x}) 25 6^{x}

\frac{d}{d x} (25 6^{x}) = ln (2) \cdot 2^{8 x + 3}

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

= ln (2) \cdot 2^{8 x + 3} x^{4 x} + 4 x^{4 x} (ln (x) + 1) \cdot 25 6^{x}

4 x^{4 x} (ln (x) + 1) \cdot 25 6^{x} = 2^{2 + 8 x} x^{4 x} (ln (x) + 1)

= ln (2) \cdot 2^{8 x + 3} x^{4 x} + 2^{2 + 8 x} x^{4 x} (ln (x) + 1)

\frac{\partial}{\partial x} (sin (5 x) cos (5 y))

\frac{d}{d x} (\frac{x ^{3}}{cos ( x )})

\frac{e ^{(- 2 s)}}{s}

\int \frac{1}{25 e ^{- 7 x} + e ^{7 x}} d x

\frac{\partial ^{2}}{\partial x \partial y} (ln (x^{2} + y^{3}))

What is the derivative of (4x^{4x}) ?
The derivative of (4x^{4x}) is ln(2)*2^{8x+3}x^{4x}+2^{2+8x}x^{4x}(ln(x)+1)
What is the first derivative of (4x^{4x}) ?
The first derivative of (4x^{4x}) is ln(2)*2^{8x+3}x^{4x}+2^{2+8x}x^{4x}(ln(x)+1)