What is the tangent of f(x)=x^x,\at x=1 ?

The tangent of f(x)=x^x,\at x=1 is y=x

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tangent of f(x)=x^x,\at x=1

tangent of

f (x) = x^{x}, at x = 1

y = x

Solution steps

Find the tangent point:

(1, 1)

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 = 1 : m = 1

Find the line with slope m=

1

and passing through

(1, 1) : y = x

y = x

\frac{\partial}{\partial s} (s e^{t})

\frac{d}{d x} ((x - 2)^{3} + 1)

\frac{\partial}{\partial x} (3 x^{3} y^{2})

\frac{3 x}{2 ( 1 - 2 x )}

\int cos (\frac{θ}{2}) d θ

What is the tangent of f(x)=x^x,\at x=1 ?
The tangent of f(x)=x^x,\at x=1 is y=x