What is the tangent of 2x^3+36x^2+210x+4 ?

The tangent of 2x^3+36x^2+210x+4 is y=(6a_{0}^2+72a_{0}+210)x-4a_{0}^3-36a_{0}^2+4

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tangent of 2x^3+36x^2+210x+4

tangent of

2 x^{3} + 36 x^{2} + 210 x + 4

y = (6 a_{0}^{2} + 72 a_{0} + 210) x - 4 a_{0}^{3} - 36 a_{0}^{2} + 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} + 36 a_{0}^{2} + 210 a_{0} + 4)

Find the slope of

y = 2 x^{3} + 36 x^{2} + 210 x + 4 : \frac{d y}{d x} = 6 x^{2} + 72 x + 210

Compute the slope of

y = 2 x^{3} + 36 x^{2} + 210 x + 4

x = a_{0} : m = 6 a_{0}^{2} + 72 a_{0} + 210

Find the line with slope m=

6 a_{0}^{2} + 72 a_{0} + 210

and passing through

(a_{0}, 2 a_{0}^{3} + 36 a_{0}^{2} + 210 a_{0} + 4) : y = (6 a_{0}^{2} + 72 a_{0} + 210) x - 4 a_{0}^{3} - 36 a_{0}^{2} + 4

y = (6 a_{0}^{2} + 72 a_{0} + 210) x - 4 a_{0}^{3} - 36 a_{0}^{2} + 4

\frac{\partial}{\partial x} (\frac{yz}{x})

\int sin (2 x) sin (2 x) d x

\int 9 - (x - 4)^{2} d x

\int \frac{( lo g _{7} ( x ) )}{x} d x

\frac{\partial}{\partial t} (\frac{c}{x + c t})

What is the tangent of 2x^3+36x^2+210x+4 ?
The tangent of 2x^3+36x^2+210x+4 is y=(6a_{0}^2+72a_{0}+210)x-4a_{0}^3-36a_{0}^2+4