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

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

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

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = 3 + 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}, 3 + 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} + 3

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

x \to 0 lim ((5 x + 2)^{3})

ln (25 sin^{2} (x))

x \to \infty lim (\frac{( 3 x ^{3} + 12 x ^{2} - 8 x + 1 )}{( x ^{2} + 2 ) ( 3 x - 1 )})

y^{^{'''}} - 9 y^{^{''}} + 27 y^{^{'}} - 27 y = 0

\frac{d}{d x} (arcsin (8 x))

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