What is the tangent of y=3e^x ?

The tangent of y=3e^x is y=3e^{a_{0}}x+3e^{a_{0}}-3e^{a_{0}}a_{0}

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tangent of y=3e^x

tangent of

y = 3 e^{x}

y = 3 e^{a_{0}} x + 3 e^{a_{0}} - 3 e^{a_{0}} a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 e^{a_{0}})

Find the slope of

y = 3 e^{x} : \frac{d y}{d x} = 3 e^{x}

Compute the slope of

y = 3 e^{x}

x = a_{0} : m = 3 e^{a_{0}}

Find the line with slope m=

3 e^{a_{0}}

and passing through

(a_{0}, 3 e^{a_{0}}) : y = 3 e^{a_{0}} x + 3 e^{a_{0}} - 3 e^{a_{0}} a_{0}

y = 3 e^{a_{0}} x + 3 e^{a_{0}} - 3 e^{a_{0}} a_{0}

f (x) = x^{3} - 3 x^{2} + 1, at x = 2

\int \frac{ln ( ln ( x ) )}{x} d x

\int_{0}^{5 π} x^{3} sin (x) d x

π r^{2} h

\int x^{3} - 4 x d x

What is the tangent of y=3e^x ?
The tangent of y=3e^x is y=3e^{a_{0}}x+3e^{a_{0}}-3e^{a_{0}}a_{0}