What is the tangent of (x^2-3)(x^3-2x) ?

The tangent of (x^2-3)(x^3-2x) is y=(5a_{0}^4-15a_{0}^2+6)x-4a_{0}^5+10a_{0}^3

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tangent of (x^2-3)(x^3-2x)

tangent of

(x^{2} - 3) (x^{3} - 2 x)

y = (5 a_{0}^{4} - 15 a_{0}^{2} + 6) x - 4 a_{0}^{5} + 10 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = (x^{2} - 3) (x^{3} - 2 x) : \frac{d y}{d x} = 5 x^{4} - 15 x^{2} + 6

Compute the slope of

y = (x^{2} - 3) (x^{3} - 2 x)

x = a_{0} : m = 5 a_{0}^{4} - 15 a_{0}^{2} + 6

Find the line with slope m=

5 a_{0}^{4} - 15 a_{0}^{2} + 6

and passing through

(a_{0}, (a_{0}^{2} - 3) (a_{0}^{3} - 2 a_{0})) : y = (5 a_{0}^{4} - 15 a_{0}^{2} + 6) x - 4 a_{0}^{5} + 10 a_{0}^{3}

y = (5 a_{0}^{4} - 15 a_{0}^{2} + 6) x - 4 a_{0}^{5} + 10 a_{0}^{3}

\int_{0}^{\frac{π}{4}} (cos (2 x)) d x

y^{^{''}} + 2 y^{^{'}} + 5 y = - 2 t + 4 t^{2}

\frac{d}{d x} (\frac{3}{x ^{3}} - \frac{1}{x ^{4}})

\frac{d}{d x} (\frac{( x ^{2} - 3 x )}{( 2 x + 1 )})

3 y^{^{''}} + y = 0

What is the tangent of (x^2-3)(x^3-2x) ?
The tangent of (x^2-3)(x^3-2x) is y=(5a_{0}^4-15a_{0}^2+6)x-4a_{0}^5+10a_{0}^3