What is the tangent of e^x+1 ?

The tangent of e^x+1 is y=e^{a_{0}}x+e^{a_{0}}+1-e^{a_{0}}a_{0}

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tangent of e^x+1

tangent of

e^{x} + 1

y = e^{a_{0}} x + e^{a_{0}} + 1 - e^{a_{0}} a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, e^{a_{0}} + 1)

Find the slope of

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

Compute the slope of

y = e^{x} + 1

x = a_{0} : m = e^{a_{0}}

Find the line with slope m=

e^{a_{0}}

and passing through

(a_{0}, e^{a_{0}} + 1) : y = e^{a_{0}} x + e^{a_{0}} + 1 - e^{a_{0}} a_{0}

y = e^{a_{0}} x + e^{a_{0}} + 1 - e^{a_{0}} a_{0}

(e^{5 x})^{^{'}}

x \to 2 + lim ((x - 3)^{2})

\int \frac{x ^{2} - 5 x + 6}{( 2 x - 1 ) ( x - 2 ) ^{2}} d x

x \to \infty lim (x^{3} - 3)

lo g_{2} (n)

What is the tangent of e^x+1 ?
The tangent of e^x+1 is y=e^{a_{0}}x+e^{a_{0}}+1-e^{a_{0}}a_{0}