What is the tangent of f(x)=e^{-x}ln(x),(1,0) ?

The tangent of f(x)=e^{-x}ln(x),(1,0) is y= 1/e x-1/e

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tangent of f(x)=e^{-x}ln(x),(1,0)

tangent of

f (x) = e^{- x} ln (x), (1, 0)

y = \frac{1}{e} x - \frac{1}{e}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = e^{- x} ln (x)

(1, 0) : m = \frac{1}{e}

Find the line with slope m=

\frac{1}{e}

and passing through

(1, 0) : y = \frac{1}{e} x - \frac{1}{e}

y = \frac{1}{e} x - \frac{1}{e}

(- 9, 0), (2, 5)

8.4

y = 2 x - 3

y = - \frac{2}{3} x + 4

(23, \frac{2 π}{3})

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