What is the tangent of y=1.8e^{2x}-41 ?

The tangent of y=1.8e^{2x}-41 is y=3.6e^{2a_{0}}x+(9e^{2a_{0}}-205-18a_{0}e^{2a_{0}})/5

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tangent of y=1.8e^{2x}-41

tangent of

y = 1.8 e^{2 x} - 41

y = 3.6 e^{2 a_{0}} x + \frac{9 e ^{2 a_{0}} - 205 - 18 a _{0} e ^{2 a_{0}}}{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}} - 41)

Find the slope of

y = 1.8 e^{2 x} - 41 : \frac{d y}{d x} = 3.6 e^{2 x}

Compute the slope of

y = 1.8 e^{2 x} - 41

x = a_{0} : 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}} - 41) : y = 3.6 e^{2 a_{0}} x + \frac{9 e ^{2 a_{0}} - 205 - 18 a _{0} e ^{2 a_{0}}}{5}

y = 3.6 e^{2 a_{0}} x + \frac{9 e ^{2 a_{0}} - 205 - 18 a _{0} e ^{2 a_{0}}}{5}

\frac{\partial}{\partial x} (x e^{y} - x^{5} + y - 3)

\frac{d}{d x} (sinh (3 x^{2} + 1))

\frac{d y}{d x} = 5 y

\frac{2}{( x - 1 ) ^{3}}

x^{4}, 8 x

What is the tangent of y=1.8e^{2x}-41 ?
The tangent of y=1.8e^{2x}-41 is y=3.6e^{2a_{0}}x+(9e^{2a_{0}}-205-18a_{0}e^{2a_{0}})/5