What is the tangent of y=x^3+6x ?

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

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tangent of y=x^3+6x

tangent of

y = x^{3} + 6 x

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} + 6 a_{0})

Find the slope of

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

Compute the slope of

y = x^{3} + 6 x

x = a_{0} : m = 3 a_{0}^{2} + 6

Find the line with slope m=

3 a_{0}^{2} + 6

and passing through

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

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

4 + x^{6}

f (x) = 2 x 4 - x^{2}

x \to 2 lim (- x^{2} + 6 x - 4)

y = 6 x - 2, (1, 3)

f (x) = \frac{2 x}{x + 1}

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