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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

(1, 0)

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)

x = 1 : 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}

x = - 2

f (x) = 4 x + 7, at x = 5

(2, 2)

tan (x)

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

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