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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derivative of

(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)

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)