What is the tangent of 5+4x^2-2x^3 ?

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

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tangent of 5+4x^2-2x^3

tangent of

5 + 4 x^{2} - 2 x^{3}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 + 4 a_{0}^{2} - 2 a_{0}^{3})

Find the slope of

y = 5 + 4 x^{2} - 2 x^{3} : \frac{d y}{d x} = - 6 x^{2} + 8 x

Compute the slope of

y = 5 + 4 x^{2} - 2 x^{3}

x = a_{0} : m = - 6 a_{0}^{2} + 8 a_{0}

Find the line with slope m=

- 6 a_{0}^{2} + 8 a_{0}

and passing through

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

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

\frac{x}{8} - \frac{8}{x}

y^{^{''}} - 7 y^{^{'}} + 6 y = 0

\int \frac{y ^{3}}{y ^{2} + 1} d y

\frac{d}{d x} (a x cos (3 x) + b x sin (3 x))

\int \frac{2}{t ^{2}} d t

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