What is the tangent of y=8+5x^2-2x^3 ?

The tangent of y=8+5x^2-2x^3 is y=(-6a_{0}^2+10a_{0})x+4a_{0}^3-5a_{0}^2+8

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tangent of y=8+5x^2-2x^3

tangent of

y = 8 + 5 x^{2} - 2 x^{3}

y = (- 6 a_{0}^{2} + 10 a_{0}) x + 4 a_{0}^{3} - 5 a_{0}^{2} + 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 + 5 a_{0}^{2} - 2 a_{0}^{3})

Find the slope of

y = 8 + 5 x^{2} - 2 x^{3} : \frac{d y}{d x} = - 6 x^{2} + 10 x

Compute the slope of

y = 8 + 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}, 8 + 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} + 8

y = (- 6 a_{0}^{2} + 10 a_{0}) x + 4 a_{0}^{3} - 5 a_{0}^{2} + 8

\frac{4 s}{( s + 2 ) ^{2}}

\frac{d}{d y} (\frac{x}{y})

\int_{0}^{\frac{π}{12}} tan (6 x) d x

\frac{x + 6}{x ^{3} + x - 7}

\int x (x + 4)^{6} d x

What is the tangent of y=8+5x^2-2x^3 ?
The tangent of y=8+5x^2-2x^3 is y=(-6a_{0}^2+10a_{0})x+4a_{0}^3-5a_{0}^2+8