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

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

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

tangent of

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

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

Solution steps

Find the slope of

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

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 = \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}

(- 6, 0)

3 x^{3}

2 + x = 6

f (x) = x^{8} 5 - 3 x

x 4 - x^{2}

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