What is the tangent of f(x)=x^x,\at x=2 ?

The tangent of f(x)=x^x,\at x=2 is y=4(ln(2)+1)x-8ln(2)-4

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tangent of f(x)=x^x,\at x=2

tangent of

f (x) = x^{x}, at x = 2

y = 4 (ln (2) + 1) x - 8 ln (2) - 4

Solution steps

Find the tangent point:

(2, 4)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 4 (ln (2) + 1)

Find the line with slope m=

4 (ln (2) + 1)

and passing through

(2, 4) : y = 4 (ln (2) + 1) x - 8 ln (2) - 4

y = 4 (ln (2) + 1) x - 8 ln (2) - 4

(x y^{2} - x) d x + (x^{2} y + y) d y = 0

\frac{\partial}{\partial x} (20 - 2 x + 1.5 - x + 3 x)

\frac{d ^{2}}{d x ^{2}} (- 8 sin (\frac{3 x}{2} + 1))

\frac{\partial}{\partial t} (x e^{- t} sin (θ))

x \to 0 + lim (\frac{1}{x ^{2}} + ln (x))

What is the tangent of f(x)=x^x,\at x=2 ?
The tangent of f(x)=x^x,\at x=2 is y=4(ln(2)+1)x-8ln(2)-4