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

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

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

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

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derivative of

f (x) = x^{x^{x}}

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

\frac{d}{d 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})

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