What is the derivative of y=x^{2x} ?

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

What is the first derivative of y=x^{2x} ?

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

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derivative of

y = x^{2 x}

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

Solution steps

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

x^{2 x} = e^{2 x l n (x)}

= \frac{d}{d x} (e^{2 x l n (x)})

Apply the chain rule:

e^{2 x l n (x)} \frac{d}{d x} (2 x ln (x))

= e^{2 x l n (x)} \frac{d}{d x} (2 x ln (x))

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

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

e^{2 x l n (x)} = x^{2 x}

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

\frac{d}{d x} (2 + 2 cos (x))

y = - x^{3} + 2, y = x - 3, x = - 1, x = 1

\int \frac{x}{y} d y

\frac{\partial}{\partial x} (ln (1 + x^{2} + y))

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