What is the derivative of (6x^{2x}) ?

The derivative of (6x^{2x}) is ln(6)*9^x*2^{2x+1}x^{2x}+9^x*2^{1+2x}x^{2x}(ln(x)+1)

What is the first derivative of (6x^{2x}) ?

The first derivative of (6x^{2x}) is ln(6)*9^x*2^{2x+1}x^{2x}+9^x*2^{1+2x}x^{2x}(ln(x)+1)

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

ln (6) \cdot 9^{x} \cdot 2^{2 x + 1} x^{2 x} + 9^{x} \cdot 2^{1 + 2 x} x^{2 x} (ln (x) + 1)

Solution steps

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

Simplify

(6 x)^{2 x} : 3 6^{x} x^{2 x}

= \frac{d}{d x} (3 6^{x} x^{2 x})

Apply the Product Rule:

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

f = 3 6^{x}, g = x^{2 x}

= \frac{d}{d x} (3 6^{x}) x^{2 x} + \frac{d}{d x} (x^{2 x}) 3 6^{x}

\frac{d}{d x} (3 6^{x}) = ln (6) \cdot 9^{x} \cdot 2^{2 x + 1}

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

= ln (6) \cdot 9^{x} \cdot 2^{2 x + 1} x^{2 x} + 2 x^{2 x} (ln (x) + 1) \cdot 3 6^{x}

2 x^{2 x} (ln (x) + 1) \cdot 3 6^{x} = 9^{x} \cdot 2^{1 + 2 x} x^{2 x} (ln (x) + 1)

= ln (6) \cdot 9^{x} \cdot 2^{2 x + 1} x^{2 x} + 9^{x} \cdot 2^{1 + 2 x} x^{2 x} (ln (x) + 1)

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\int (1 - cos (\frac{x}{2}))^{2} sin (\frac{x}{2}) d x

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What is the derivative of (6x^{2x}) ?
The derivative of (6x^{2x}) is ln(6)*9^x*2^{2x+1}x^{2x}+9^x*2^{1+2x}x^{2x}(ln(x)+1)
What is the first derivative of (6x^{2x}) ?
The first derivative of (6x^{2x}) is ln(6)*9^x*2^{2x+1}x^{2x}+9^x*2^{1+2x}x^{2x}(ln(x)+1)