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

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

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

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

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

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

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

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

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

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

y^{^{''}} - (\frac{2}{t}) y^{^{'}} + (\frac{2}{t ^{2}}) y = \frac{2 t}{1 + t ^{2}}

\int x^{9} x^{5} + 4 d x

\frac{d}{d y} (y^{- 2})

x - 1

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

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