What is the tangent of-6x^2+9x+4x^3 ?

The tangent of-6x^2+9x+4x^3 is y=(-12a_{0}+9+12a_{0}^2)x-8a_{0}^3+6a_{0}^2

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tangent of-6x^2+9x+4x^3

tangent of

- 6 x^{2} + 9 x + 4 x^{3}

y = (- 12 a_{0} + 9 + 12 a_{0}^{2}) x - 8 a_{0}^{3} + 6 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 6 a_{0}^{2} + 9 a_{0} + 4 a_{0}^{3})

Find the slope of

y = - 6 x^{2} + 9 x + 4 x^{3} : \frac{d y}{d x} = - 12 x + 9 + 12 x^{2}

Compute the slope of

y = - 6 x^{2} + 9 x + 4 x^{3}

x = a_{0} : m = - 12 a_{0} + 9 + 12 a_{0}^{2}

Find the line with slope m=

- 12 a_{0} + 9 + 12 a_{0}^{2}

and passing through

(a_{0}, - 6 a_{0}^{2} + 9 a_{0} + 4 a_{0}^{3}) : y = (- 12 a_{0} + 9 + 12 a_{0}^{2}) x - 8 a_{0}^{3} + 6 a_{0}^{2}

y = (- 12 a_{0} + 9 + 12 a_{0}^{2}) x - 8 a_{0}^{3} + 6 a_{0}^{2}

\frac{d}{d x} ((8 x + 1)^{2})

\int \frac{17}{( x + 1 ) ( x ^{2} + 16 )} d x

\int sec^{3} (θ) d θ

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

\frac{d}{d x} ((sin^{2} (x)) d x)

What is the tangent of-6x^2+9x+4x^3 ?
The tangent of-6x^2+9x+4x^3 is y=(-12a_{0}+9+12a_{0}^2)x-8a_{0}^3+6a_{0}^2