What is the tangent of 10x^3-30x+120 ?

The tangent of 10x^3-30x+120 is y=(30a_{0}^2-30)x-20a_{0}^3+120

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tangent of 10x^3-30x+120

tangent of

10 x^{3} - 30 x + 120

y = (30 a_{0}^{2} - 30) x - 20 a_{0}^{3} + 120

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 10 a_{0}^{3} - 30 a_{0} + 120)

Find the slope of

y = 10 x^{3} - 30 x + 120 : \frac{d y}{d x} = 30 x^{2} - 30

Compute the slope of

y = 10 x^{3} - 30 x + 120

x = a_{0} : m = 30 a_{0}^{2} - 30

Find the line with slope m=

30 a_{0}^{2} - 30

and passing through

(a_{0}, 10 a_{0}^{3} - 30 a_{0} + 120) : y = (30 a_{0}^{2} - 30) x - 20 a_{0}^{3} + 120

y = (30 a_{0}^{2} - 30) x - 20 a_{0}^{3} + 120

\frac{d}{d x} (2 x^{2} y^{2})

z + 3 (\frac{1}{z ^{2}})

\frac{d y}{d x} + y tan (x) = 2 x cos (x)

\frac{\partial}{\partial z} (x y + yz + x z)

\int \frac{1}{3} x^{3} d x

What is the tangent of 10x^3-30x+120 ?
The tangent of 10x^3-30x+120 is y=(30a_{0}^2-30)x-20a_{0}^3+120