What is the (x^{x^{x^x}})^' ?

The (x^{x^{x^x}})^' is x^{x^{x^x}}(x^{x^x}ln(x)(x^xln(x)(ln(x)+1)+x^{x-1})+x^{x^x-1})

What is the first (x^{x^{x^x}})^' ?

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

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

(7 x + 6 y) d x + (6 x - 8 y 3) d y = 0

x \to 0 lim (\frac{sin ( 7 x )}{x})

\frac{d}{d x} (1 - tan (x))

\frac{d ^{2}}{d x ^{2}} (cos^{2} (4 π x))

y^{^{'''}} - y^{^{''}} - y^{^{'}} = 0

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