What is the tangent of-x^3(-2.8) ?

The tangent of-x^3(-2.8) is y=8.4a_{0}^2x-(28a_{0}^3)/5

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tangent of-x^3(-2.8)

tangent of

- x^{3} (- 2.8)

y = 8.4 a_{0}^{2} x - \frac{28 a _{0}^{3}}{5}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2.8 a_{0}^{3})

Find the slope of

y = - x^{3} (- 2.8) : \frac{d y}{d x} = 8.4 x^{2}

Compute the slope of

y = - x^{3} (- 2.8)

x = a_{0} : m = 8.4 a_{0}^{2}

Find the line with slope m=

8.4 a_{0}^{2}

and passing through

(a_{0}, 2.8 a_{0}^{3}) : y = 8.4 a_{0}^{2} x - \frac{28 a _{0}^{3}}{5}

y = 8.4 a_{0}^{2} x - \frac{28 a _{0}^{3}}{5}

\int \frac{x ^{2}}{e ^{2 x}} d x

\frac{d}{d θ} (7 sin (θ))

- 2 x sin (x^{2})

\frac{d}{d x} ((sin (x))^{3})

\frac{1}{4} y^{^{''}} + 4 y = tan (4 t) - \frac{1}{2} e^{3 t}

What is the tangent of-x^3(-2.8) ?
The tangent of-x^3(-2.8) is y=8.4a_{0}^2x-(28a_{0}^3)/5