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

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

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

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

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derivative of

f (x) = 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)

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