What is the derivative of (8x)^{4x} ?

The derivative of (8x)^{4x} is 3ln(2)*2^{12x+2}x^{4x}+2^{2+12x}x^{4x}(ln(x)+1)

What is the first derivative of (8x)^{4x} ?

The first derivative of (8x)^{4x} is 3ln(2)*2^{12x+2}x^{4x}+2^{2+12x}x^{4x}(ln(x)+1)

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derivative of

(8 x)^{4 x}

3 ln (2) \cdot 2^{12 x + 2} x^{4 x} + 2^{2 + 12 x} x^{4 x} (ln (x) + 1)

Solution steps

\frac{d}{d x} ((8 x)^{4 x})

Simplify

(8 x)^{4 x} : 409 6^{x} x^{4 x}

= \frac{d}{d x} (409 6^{x} x^{4 x})

Apply the Product Rule:

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

f = 409 6^{x}, g = x^{4 x}

= \frac{d}{d x} (409 6^{x}) x^{4 x} + \frac{d}{d x} (x^{4 x}) 409 6^{x}

\frac{d}{d x} (409 6^{x}) = 3 ln (2) \cdot 2^{12 x + 2}

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

= 3 ln (2) \cdot 2^{12 x + 2} x^{4 x} + 4 x^{4 x} (ln (x) + 1) \cdot 409 6^{x}

4 x^{4 x} (ln (x) + 1) \cdot 409 6^{x} = 2^{2 + 12 x} x^{4 x} (ln (x) + 1)

= 3 ln (2) \cdot 2^{12 x + 2} x^{4 x} + 2^{2 + 12 x} x^{4 x} (ln (x) + 1)

\frac{\partial}{\partial x} (2 x sin (2 y))

\frac{d}{d x} (A x cos (x) + B cos (x))

x \to 1 lim (\frac{1}{x - 1})

\frac{1}{2 s + 1}

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

What is the derivative of (8x)^{4x} ?
The derivative of (8x)^{4x} is 3ln(2)*2^{12x+2}x^{4x}+2^{2+12x}x^{4x}(ln(x)+1)
What is the first derivative of (8x)^{4x} ?
The first derivative of (8x)^{4x} is 3ln(2)*2^{12x+2}x^{4x}+2^{2+12x}x^{4x}(ln(x)+1)