What is the tangent of y=e^xln(x),(1,0) ?

The tangent of y=e^xln(x),(1,0) is y=ex-e

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tangent of y=e^xln(x),(1,0)

tangent of

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

y = e x - 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}

Compute the slope of

y = e^{x} ln (x)

(1, 0) : m = e

Find the line with slope m=

e

and passing through

(1, 0) : y = e x - e

y = e x - e

\frac{\partial}{\partial y} (e^{x} y^{2})

\frac{d}{d x} (tan (x) - 1)

(2.4) (0.8)

\frac{d r}{d t} = \frac{sec ^{2} ( t )}{tan ( t ) + 1}

\int (\frac{9}{x} + \frac{6}{x ^{4}}) d x

What is the tangent of y=e^xln(x),(1,0) ?
The tangent of y=e^xln(x),(1,0) is y=ex-e