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}

\int \frac{1}{x ( x ^{2} + 4 ) ^{2}} d x

\frac{d}{d θ} (2 sin (5 θ))

6^{x^{3}}

\int_{1}^{10} \frac{x}{x ^{2} - 4} d x

g (x) = 2^{x^{2} + 1 x}

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