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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

Simplify

e^{x l n (x^{2})} (ln (x^{2}) + 2) : x^{2 x} (2 ln (x) + 2)

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

\int (2 y) d y

\int \frac{( 5 x - 7 )}{( x ^{2} - x - 2 ) ( x - 3 )} d x

y^{^{'}} + x y = - x^{2}

y = (5 x - 4)^{5} (3 - x^{3})^{5}

f (x) = 2 x^{2} - 6 x, at x = 1

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