What is the tangent of 9+5x^2-2x^3 ?

The tangent of 9+5x^2-2x^3 is y=(-6a_{0}^2+10a_{0})x+4a_{0}^3-5a_{0}^2+9

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tangent of 9+5x^2-2x^3

tangent of

9 + 5 x^{2} - 2 x^{3}

y = (- 6 a_{0}^{2} + 10 a_{0}) x + 4 a_{0}^{3} - 5 a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 9 + 5 a_{0}^{2} - 2 a_{0}^{3})

Find the slope of

y = 9 + 5 x^{2} - 2 x^{3} : \frac{d y}{d x} = - 6 x^{2} + 10 x

Compute the slope of

y = 9 + 5 x^{2} - 2 x^{3}

x = a_{0} : m = - 6 a_{0}^{2} + 10 a_{0}

Find the line with slope m=

- 6 a_{0}^{2} + 10 a_{0}

and passing through

(a_{0}, 9 + 5 a_{0}^{2} - 2 a_{0}^{3}) : y = (- 6 a_{0}^{2} + 10 a_{0}) x + 4 a_{0}^{3} - 5 a_{0}^{2} + 9

y = (- 6 a_{0}^{2} + 10 a_{0}) x + 4 a_{0}^{3} - 5 a_{0}^{2} + 9

\int \frac{x ^{2} - 1}{x ^{4}} d x

16 y^{^{''}} + 7 y = 0

\frac{d}{d x} (1 - e^{x} + (e - 1) sin ((\frac{π}{2}) x))

y^{^{''}} + 2 y^{^{'}} + y = 2 e - t

x \to 0 + lim (\frac{x ^{2}}{8} - \frac{6}{x})

What is the tangent of 9+5x^2-2x^3 ?
The tangent of 9+5x^2-2x^3 is y=(-6a_{0}^2+10a_{0})x+4a_{0}^3-5a_{0}^2+9