What is the tangent of y=3-x^3(0.3) ?

The tangent of y=3-x^3(0.3) is y=-0.9a_{0}^2x+(3(a_{0}^3+5))/5

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tangent of y=3-x^3(0.3)

tangent of

y = 3 - x^{3} (0.3)

y = - 0.9 a_{0}^{2} x + \frac{3 ( a _{0}^{3} + 5 )}{5}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 - a_{0}^{3} \cdot 0.3)

Find the slope of

y = 3 - x^{3} 0.3 : \frac{d y}{d x} = - 0.9 x^{2}

Compute the slope of

y = 3 - x^{3} 0.3

x = a_{0} : m = - 0.9 a_{0}^{2}

Find the line with slope m=

- 0.9 a_{0}^{2}

and passing through

(a_{0}, 3 - a_{0}^{3} 0.3) : y = - 0.9 a_{0}^{2} x + \frac{3 ( a _{0}^{3} + 5 )}{5}

y = - 0.9 a_{0}^{2} x + \frac{3 ( a _{0}^{3} + 5 )}{5}

x \to \infty lim (1)

\frac{d}{d x} (lo g_{10} (2 x - 1))

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

\frac{d}{d x} ((5 x + 2)^{4} (2 x - 3)^{8})

y = 4 x, y = 0, x = 2

What is the tangent of y=3-x^3(0.3) ?
The tangent of y=3-x^3(0.3) is y=-0.9a_{0}^2x+(3(a_{0}^3+5))/5