What is the tangent of e^x,ln(12) ?

The tangent of e^x,ln(12) is y=e^{a_{0}}x+e^{a_{0}}-e^{a_{0}}a_{0}

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tangent of e^x,ln(12)

tangent of

e^{x}, ln (12)

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

Find the slope of

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

Compute the slope of

y = e^{x}

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

Find the line with slope m=

e^{a_{0}}

and passing through

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

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

\frac{d}{d x} (x^{2} sin (\frac{1}{x ^{2}}))

x \to 0 lim (\frac{9}{x})

\frac{d y}{d x} = - \frac{1 + x + y}{1 + x}

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 0

\int 4 + x d x

What is the tangent of e^x,ln(12) ?
The tangent of e^x,ln(12) is y=e^{a_{0}}x+e^{a_{0}}-e^{a_{0}}a_{0}