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

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

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

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

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

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

Solution steps

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

Simplify

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

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

Apply the Product Rule:

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

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

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

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

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

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

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

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

\frac{d}{d x} (4 x^{3} sec (7 x))

(x^{2} + 1) y^{^{'}} + 7 x (y - 1) = 0

(x - 2) y^{^{'}} - y = 2 (x - 2)^{3}, y (3) = 5

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

x^{2} + 16, at x = 3

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