What is the tangent of f(x)=[1+e^x][ln(x)],\at x=1 ?

The tangent of f(x)=[1+e^x][ln(x)],\at x=1 is y=(e+1)x-e-1

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tangent of f(x)=[1+e^x][ln(x)],\at x=1

tangent of

f (x) = [1 + e^{x}] [ln (x)], at x = 1

y = (e + 1) x - e - 1

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = [1 + e^{x}] [ln (x)] : \frac{df ( x )}{d x} = e^{x} ln (x) + \frac{1 + e ^{x}}{x}

Compute the slope of

f (x) = [1 + e^{x}] [ln (x)]

x = 1 : m = e + 1

Find the line with slope m=

e + 1

and passing through

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

y = (e + 1) x - e - 1

f (x) = e^{x} cos (x)

v^{''} - v = 0

\int t e^{- s t} sin (t) d t

\frac{\partial}{\partial x} (- \frac{x cos ( y )}{x - 1})

sin (tan (6 x))

What is the tangent of f(x)=[1+e^x][ln(x)],\at x=1 ?
The tangent of f(x)=[1+e^x][ln(x)],\at x=1 is y=(e+1)x-e-1