What is the tangent of 2x^3(2.16) ?

The tangent of 2x^3(2.16) is y=12.96a_{0}^2x-(216a_{0}^3)/(25)

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tangent of 2x^3(2.16)

tangent of

2 x^{3} (2.16)

y = 12.96 a_{0}^{2} x - \frac{216 a _{0}^{3}}{25}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4.32 a_{0}^{3})

Find the slope of

y = 2 x^{3} 2.16 : \frac{d y}{d x} = 12.96 x^{2}

Compute the slope of

y = 2 x^{3} 2.16

x = a_{0} : m = 12.96 a_{0}^{2}

Find the line with slope m=

12.96 a_{0}^{2}

and passing through

(a_{0}, 4.32 a_{0}^{3}) : y = 12.96 a_{0}^{2} x - \frac{216 a _{0}^{3}}{25}

y = 12.96 a_{0}^{2} x - \frac{216 a _{0}^{3}}{25}

\frac{d}{d x} (120)

x \to - 0 lim (\frac{1}{x})

y = x, y = x^{2} + 1, x = - 1, x = 2

y^{^{'}} = \frac{1}{y ^{2}}

y = cos (x), at x = 2 π

What is the tangent of 2x^3(2.16) ?
The tangent of 2x^3(2.16) is y=12.96a_{0}^2x-(216a_{0}^3)/(25)