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

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

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

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

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derivative of

y = 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})

(x^{5} + 3 y) d x - x d y = 0

\int \frac{x ^{3}}{49 + x ^{8}} d x

\frac{d ^{2}}{d x ^{2}} (15 e^{x^{2}})

9 - x^{2}

\frac{\partial}{\partial x} ((2 x + 6 y)^{11})

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