What is the tangent of y=x+e^x-1/x ,\at x=1 ?

The tangent of y=x+e^x-1/x ,\at x=1 is y=(e+2)x-2

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tangent of y=x+e^x-1/x ,\at x=1

tangent of

y = x + e^{x} - \frac{1}{x}, at x = 1

y = (e + 2) x - 2

Solution steps

Find the tangent point:

(1, e)

Find the slope of

y = x + e^{x} - \frac{1}{x} : \frac{d y}{d x} = 1 + e^{x} + \frac{1}{x ^{2}}

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 = e + 2

Find the line with slope m=

e + 2

and passing through

(1, e) : y = (e + 2) x - 2

y = (e + 2) x - 2

(- 3 x^{2} - 3)^{4}

x \to 2 lim (x^{2} + x + 2)

\frac{\partial}{\partial x} (x^{2} y z^{2} + sin (yz))

\frac{d}{d x} (2 + sin (3 x))

\int sin (x^{2} + y^{2})

What is the tangent of y=x+e^x-1/x ,\at x=1 ?
The tangent of y=x+e^x-1/x ,\at x=1 is y=(e+2)x-2