What is the tangent of-3-3x^2 ?

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

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tangent of-3-3x^2

tangent of

- 3 - 3 x^{2}

y = - 6 a_{0} x - 3 + 3 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = - 3 - 3 x^{2}

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

Find the line with slope m=

- 6 a_{0}

and passing through

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

y = - 6 a_{0} x - 3 + 3 a_{0}^{2}

f (x) = \frac{3 - 2 x - x ^{2}}{x ^{2} - 1}

\frac{d y}{d x} = x (1 + y)

\frac{\partial}{\partial x} (2 x y + y^{2} + 1)

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

y^{2} d x - x y d y = x^{2} y d y