What is the derivative of (8x)^{2x} ?

The derivative of (8x)^{2x} is 3ln(2)*2^{6x+1}x^{2x}+2^{1+6x}x^{2x}(ln(x)+1)

What is the first derivative of (8x)^{2x} ?

The first derivative of (8x)^{2x} is 3ln(2)*2^{6x+1}x^{2x}+2^{1+6x}x^{2x}(ln(x)+1)

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derivative of

(8 x)^{2 x}

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

Solution steps

\frac{d}{d x} ((8 x)^{2 x})

Simplify

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

= \frac{d}{d x} (6 4^{x} x^{2 x})

Apply the Product Rule:

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

f = 6 4^{x}, g = x^{2 x}

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

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

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

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

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

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

\frac{\partial}{\partial s} (\frac{s - 2 r}{t})

\int \frac{3 x + 1}{x} d x

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = 0

\frac{d}{d x} (2 x^{2} + 7)

f (x) = (4 x + 5) (5 x - 1), at x = 3

What is the derivative of (8x)^{2x} ?
The derivative of (8x)^{2x} is 3ln(2)*2^{6x+1}x^{2x}+2^{1+6x}x^{2x}(ln(x)+1)
What is the first derivative of (8x)^{2x} ?
The first derivative of (8x)^{2x} is 3ln(2)*2^{6x+1}x^{2x}+2^{1+6x}x^{2x}(ln(x)+1)