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Popular >

(x^{x^x})^'

Solution

(x^{x^{x}})^{^{'}}

Solution

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

Solution steps

(x^{x^{x}})^{'}

Apply exponent rule:

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

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

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

Apply the chain rule:

e^{x^{x} l n (x)} (x^{x} ln (x))^{'}

= e^{x^{x} l n (x)} (x^{x} ln (x))^{'}

(x^{x} ln (x))^{'} = x^{x} ln (x) (ln (x) + 1) + x^{x - 1}

= e^{x^{x} l n (x)} (x^{x} ln (x) (ln (x) + 1) + x^{x - 1})

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

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

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Frequently Asked Questions (FAQ)

What is the (x^{x^x})^' ?
The (x^{x^x})^' is x^{x^x}(x^xln(x)(ln(x)+1)+x^{x-1})
What is the first (x^{x^x})^' ?
The first (x^{x^x})^' is x^{x^x}(x^xln(x)(ln(x)+1)+x^{x-1})