What is the tangent of f(x)=x^x ?

The tangent of f(x)=x^x is y=a_{0}^{a_{0}}(ln(a_{0})+1)x+a_{0}^{a_{0}}-a_{0}^{a_{0}+1}(ln(a_{0})+1)

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tangent of f(x)=x^x

tangent of

f (x) = x^{x}

y = a_{0}^{a_{0}} (ln (a_{0}) + 1) x + a_{0}^{a_{0}} - a_{0}^{a_{0} + 1} (ln (a_{0}) + 1)

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{a_{0}})

Find the slope of

f (x) = x^{x} : \frac{df ( x )}{d x} = x^{x} (ln (x) + 1)

Compute the slope of

f (x) = x^{x}

x = a_{0} : m = a_{0}^{a_{0}} (ln (a_{0}) + 1)

Find the line with slope m=

a_{0}^{a_{0}} (ln (a_{0}) + 1)

and passing through

(a_{0}, a_{0}^{a_{0}}) : y = a_{0}^{a_{0}} (ln (a_{0}) + 1) x + a_{0}^{a_{0}} - a_{0}^{a_{0} + 1} (ln (a_{0}) + 1)

y = a_{0}^{a_{0}} (ln (a_{0}) + 1) x + a_{0}^{a_{0}} - a_{0}^{a_{0} + 1} (ln (a_{0}) + 1)

\frac{d}{d x} (31 arctan (x))

\frac{\partial}{\partial x} (e^{y} + x + x^{2} + 4)

x \to 2 - lim ((x - 3)^{2})

\int (3 u + 2)^{3} d u

5 x^{2} + 3 x - 8

What is the tangent of f(x)=x^x ?
The tangent of f(x)=x^x is y=a_{0}^{a_{0}}(ln(a_{0})+1)x+a_{0}^{a_{0}}-a_{0}^{a_{0}+1}(ln(a_{0})+1)