What is the tangent of 2x^3-x^2+2(1.3) ?

The tangent of 2x^3-x^2+2(1.3) is y=(6a_{0}^2-2a_{0})x-4a_{0}^3+a_{0}^2+2.6

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tangent of 2x^3-x^2+2(1.3)

tangent of

2 x^{3} - x^{2} + 2 (1.3)

y = (6 a_{0}^{2} - 2 a_{0}) x - 4 a_{0}^{3} + a_{0}^{2} + 2.6

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} + 2.6)

Find the slope of

y = 2 x^{3} - x^{2} + 2 \cdot 1.3 : \frac{d y}{d x} = 6 x^{2} - 2 x

Compute the slope of

y = 2 x^{3} - x^{2} + 2 \cdot 1.3

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} + 2.6) : y = (6 a_{0}^{2} - 2 a_{0}) x - 4 a_{0}^{3} + a_{0}^{2} + 2.6

y = (6 a_{0}^{2} - 2 a_{0}) x - 4 a_{0}^{3} + a_{0}^{2} + 2.6

\int \frac{2 x ^{2} + x - 4}{x ^{3} - x ^{2} - 2 x} d x

\frac{\partial}{\partial x} (7 sin (x) sin (y))

\frac{d}{d x} (\frac{3}{4})

\frac{d}{d x} (\frac{6}{5})

\frac{d}{d v} (\frac{v}{t})

What is the tangent of 2x^3-x^2+2(1.3) ?
The tangent of 2x^3-x^2+2(1.3) is y=(6a_{0}^2-2a_{0})x-4a_{0}^3+a_{0}^2+2.6