What is the tangent of y=1+4x^2-2x^3 ?

The tangent of y=1+4x^2-2x^3 is y=(-6a_{0}^2+8a_{0})x+4a_{0}^3-4a_{0}^2+1

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tangent of y=1+4x^2-2x^3

tangent of

y = 1 + 4 x^{2} - 2 x^{3}

y = (- 6 a_{0}^{2} + 8 a_{0}) x + 4 a_{0}^{3} - 4 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 1 + 4 a_{0}^{2} - 2 a_{0}^{3})

Find the slope of

y = 1 + 4 x^{2} - 2 x^{3} : \frac{d y}{d x} = - 6 x^{2} + 8 x

Compute the slope of

y = 1 + 4 x^{2} - 2 x^{3}

x = a_{0} : m = - 6 a_{0}^{2} + 8 a_{0}

Find the line with slope m=

- 6 a_{0}^{2} + 8 a_{0}

and passing through

(a_{0}, 1 + 4 a_{0}^{2} - 2 a_{0}^{3}) : y = (- 6 a_{0}^{2} + 8 a_{0}) x + 4 a_{0}^{3} - 4 a_{0}^{2} + 1

y = (- 6 a_{0}^{2} + 8 a_{0}) x + 4 a_{0}^{3} - 4 a_{0}^{2} + 1

\frac{\partial}{\partial y} (x^{3} - x y^{2} + 3 y^{2})

y = 4 x^{3} - 2 x, (1, 2)

f (x) = x^{2} + 1, (- 4, 17)

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

\frac{z ^{\frac{8}{3}}}{8} - \frac{e ^{- 2 z}}{2}

What is the tangent of y=1+4x^2-2x^3 ?
The tangent of y=1+4x^2-2x^3 is y=(-6a_{0}^2+8a_{0})x+4a_{0}^3-4a_{0}^2+1