What is the (x^{x^2})^' ?

The (x^{x^2})^' is x^{x^2}(2xln(x)+x)

What is the first (x^{x^2})^' ?

The first (x^{x^2})^' is x^{x^2}(2xln(x)+x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^{x^{2}})^{^{'}}

x^{x^{2}} (2 x ln (x) + x)

Solution steps

(x^{x^{2}})^{'}

Apply exponent rule:

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

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

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

Apply the chain rule:

e^{x^{2} l n (x)} (x^{2} ln (x))^{'}

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

(x^{2} ln (x))^{'} = 2 x ln (x) + x

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

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

= x^{x^{2}} (2 x ln (x) + x)

e^{- 3 \times 0}

(49)^{3}

e^{- 3 t} \cdot e^{t}

11.5 \times 1 0^{- 6}