What is the tangent of y=2x^3-x^2+7 ?

The tangent of y=2x^3-x^2+7 is y=(6a_{0}^2-2a_{0})x-4a_{0}^3+a_{0}^2+7

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tangent of y=2x^3-x^2+7

tangent of

y = 2 x^{3} - x^{2} + 7

y = (6 a_{0}^{2} - 2 a_{0}) x - 4 a_{0}^{3} + a_{0}^{2} + 7

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{3} - a_{0}^{2} + 7)

Find the slope of

y = 2 x^{3} - x^{2} + 7 : \frac{d y}{d x} = 6 x^{2} - 2 x

Compute the slope of

y = 2 x^{3} - x^{2} + 7

x = a_{0} : m = 6 a_{0}^{2} - 2 a_{0}

Find the line with slope m=

6 a_{0}^{2} - 2 a_{0}

and passing through

(a_{0}, 2 a_{0}^{3} - a_{0}^{2} + 7) : y = (6 a_{0}^{2} - 2 a_{0}) x - 4 a_{0}^{3} + a_{0}^{2} + 7

y = (6 a_{0}^{2} - 2 a_{0}) x - 4 a_{0}^{3} + a_{0}^{2} + 7

\frac{\partial}{\partial x} (\frac{t}{a ^{2} t ^{2} - x ^{2}})

\frac{e ^{x} + 1}{e ^{x} - 1}

\frac{d y}{d x} = 2 x y^{2} + 128 x

x \to - 6 lim (12 x + 1)

\frac{d y}{d x} = \frac{1}{2} (y - 1)^{2}, y (0) = 2

What is the tangent of y=2x^3-x^2+7 ?
The tangent of y=2x^3-x^2+7 is y=(6a_{0}^2-2a_{0})x-4a_{0}^3+a_{0}^2+7