What is the tangent of y=2x^3-x^2+6,(2,18) ?

The tangent of y=2x^3-x^2+6,(2,18) is y=20x-22

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tangent of y=2x^3-x^2+6,(2,18)

tangent of

y = 2 x^{3} - x^{2} + 6, (2, 18)

y = 20 x - 22

Solution steps

Find the slope of

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

Compute the slope of

y = 2 x^{3} - x^{2} + 6

(2, 18) : m = 20

Find the line with slope m=

20

and passing through

(2, 18) : y = 20 x - 22

y = 20 x - 22

y^{^{'}} = \frac{y}{x} + tan (\frac{y}{x})

4 y^{^{''}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 3

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

\frac{d}{d x} (a (2 e^{- 3 x} x - 3 e^{- 3 x} x^{2}))

\frac{\partial}{\partial x} (- sin (4 x))

What is the tangent of y=2x^3-x^2+6,(2,18) ?
The tangent of y=2x^3-x^2+6,(2,18) is y=20x-22