What is the tangent of 8e^x ?

The tangent of 8e^x is y=8e^{a_{0}}x+8e^{a_{0}}-8e^{a_{0}}a_{0}

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tangent of 8e^x

tangent of

8 e^{x}

y = 8 e^{a_{0}} x + 8 e^{a_{0}} - 8 e^{a_{0}} a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 e^{a_{0}})

Find the slope of

y = 8 e^{x} : \frac{d y}{d x} = 8 e^{x}

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 = 8 e^{a_{0}}

Find the line with slope m=

8 e^{a_{0}}

and passing through

(a_{0}, 8 e^{a_{0}}) : y = 8 e^{a_{0}} x + 8 e^{a_{0}} - 8 e^{a_{0}} a_{0}

y = 8 e^{a_{0}} x + 8 e^{a_{0}} - 8 e^{a_{0}} a_{0}

y^{^{'}} + \frac{3 y}{x} = - \frac{1}{x}

\frac{\partial}{\partial x} (2 x^{3} + x y^{2})

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

\frac{d}{d x} (\frac{1}{2 x ln ( x ) + 1})

\int cot^{3} (x) \cdot csc^{3} (x) d x

What is the tangent of 8e^x ?
The tangent of 8e^x is y=8e^{a_{0}}x+8e^{a_{0}}-8e^{a_{0}}a_{0}