What is the tangent of y=1+2x-x^3(1.2) ?

The tangent of y=1+2x-x^3(1.2) is y=(-3.6a_{0}^2+2)x+2.4a_{0}^3+1

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tangent of y=1+2x-x^3(1.2)

tangent of

y = 1 + 2 x - x^{3} (1.2)

y = (- 3.6 a_{0}^{2} + 2) x + 2.4 a_{0}^{3} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 1 + 2 a_{0} - a_{0}^{3} \cdot 1.2)

Find the slope of

y = 1 + 2 x - x^{3} 1.2 : \frac{d y}{d x} = - 3.6 x^{2} + 2

Compute the slope of

y = 1 + 2 x - x^{3} 1.2

x = a_{0} : m = - 3.6 a_{0}^{2} + 2

Find the line with slope m=

- 3.6 a_{0}^{2} + 2

and passing through

(a_{0}, 1 + 2 a_{0} - a_{0}^{3} 1.2) : y = (- 3.6 a_{0}^{2} + 2) x + 2.4 a_{0}^{3} + 1

y = (- 3.6 a_{0}^{2} + 2) x + 2.4 a_{0}^{3} + 1

\frac{d}{d x} (x^{2} ∣ x ∣)

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

\frac{d}{d x} (3 x^{2} - \frac{3}{x ^{4}})

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

x \to - 1 - lim (5 x + 2 p (x))

What is the tangent of y=1+2x-x^3(1.2) ?
The tangent of y=1+2x-x^3(1.2) is y=(-3.6a_{0}^2+2)x+2.4a_{0}^3+1