What is the tangent of 1.8e^{2x}-45 ?

The tangent of 1.8e^{2x}-45 is y=3.6e^{2a_{0}}x+(9(e^{2a_{0}}-2a_{0}e^{2a_{0}}-25))/5

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tangent of 1.8e^{2x}-45

tangent of

1.8 e^{2 x} - 45

y = 3.6 e^{2 a_{0}} x + \frac{9 ( e ^{2 a_{0}} - 2 a _{0} e ^{2 a_{0}} - 25 )}{5}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 1.8 e^{2 a_{0}} - 45)

Find the slope of

y = 1.8 e^{2 x} - 45 : \frac{d y}{d x} = 3.6 e^{2 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 = 3.6 e^{2 a_{0}}

Find the line with slope m=

3.6 e^{2 a_{0}}

and passing through

(a_{0}, 1.8 e^{2 a_{0}} - 45) : y = 3.6 e^{2 a_{0}} x + \frac{9 ( e ^{2 a_{0}} - 2 a _{0} e ^{2 a_{0}} - 25 )}{5}

y = 3.6 e^{2 a_{0}} x + \frac{9 ( e ^{2 a_{0}} - 2 a _{0} e ^{2 a_{0}} - 25 )}{5}

y = \frac{x}{6 + x}

\frac{5}{s ^{2} + 49}

n = 0 \sum \infty (- \frac{5}{4})^{- 2 n}

h \to 0 lim (\frac{a ^{h} - 1}{h})

y^{^{'}} = \frac{2 y}{tan ( x )}

What is the tangent of 1.8e^{2x}-45 ?
The tangent of 1.8e^{2x}-45 is y=3.6e^{2a_{0}}x+(9(e^{2a_{0}}-2a_{0}e^{2a_{0}}-25))/5