What is the tangent of y=6x^2-x^3(1.5) ?

The tangent of y=6x^2-x^3(1.5) is y=(12a_{0}-4.5a_{0}^2)x+3a_{0}^3-6a_{0}^2

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tangent of y=6x^2-x^3(1.5)

tangent of

y = 6 x^{2} - x^{3} (1.5)

y = (12 a_{0} - 4.5 a_{0}^{2}) x + 3 a_{0}^{3} - 6 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 6 a_{0}^{2} - a_{0}^{3} \cdot 1.5)

Find the slope of

y = 6 x^{2} - x^{3} 1.5 : \frac{d y}{d x} = 12 x - 4.5 x^{2}

Compute the slope of

y = 6 x^{2} - x^{3} 1.5

x = a_{0} : m = 12 a_{0} - 4.5 a_{0}^{2}

Find the line with slope m=

12 a_{0} - 4.5 a_{0}^{2}

and passing through

(a_{0}, 6 a_{0}^{2} - a_{0}^{3} 1.5) : y = (12 a_{0} - 4.5 a_{0}^{2}) x + 3 a_{0}^{3} - 6 a_{0}^{2}

y = (12 a_{0} - 4.5 a_{0}^{2}) x + 3 a_{0}^{3} - 6 a_{0}^{2}

x \to 0 lim (1 - x^{2})

\int \frac{x}{( 1 + x ) ^{2}} d x

x + y y^{^{'}} = 0, y (1) = 9

f (x) = 2 tan (x), at x = \frac{π}{4}

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

What is the tangent of y=6x^2-x^3(1.5) ?
The tangent of y=6x^2-x^3(1.5) is y=(12a_{0}-4.5a_{0}^2)x+3a_{0}^3-6a_{0}^2