What is the tangent of y=5x^2-x^3 ?

The tangent of y=5x^2-x^3 is y=(10a_{0}-3a_{0}^2)x+2a_{0}^3-5a_{0}^2

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tangent of y=5x^2-x^3

tangent of

y = 5 x^{2} - x^{3}

y = (10 a_{0} - 3 a_{0}^{2}) x + 2 a_{0}^{3} - 5 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 a_{0}^{2} - a_{0}^{3})

Find the slope of

y = 5 x^{2} - x^{3} : \frac{d y}{d x} = 10 x - 3 x^{2}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 10 a_{0} - 3 a_{0}^{2}

Find the line with slope m=

10 a_{0} - 3 a_{0}^{2}

and passing through

(a_{0}, 5 a_{0}^{2} - a_{0}^{3}) : y = (10 a_{0} - 3 a_{0}^{2}) x + 2 a_{0}^{3} - 5 a_{0}^{2}

y = (10 a_{0} - 3 a_{0}^{2}) x + 2 a_{0}^{3} - 5 a_{0}^{2}

x \to 2 lim (\frac{x ^{2} + 4}{x - 2})

\frac{\partial}{\partial x} (3 x^{2} \cdot e^{x^{2} y})

cos ((2 x^{2} + 1)^{7})

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

\frac{1}{2 r} + \frac{1}{5 r ^{\frac{4}{5}}}

What is the tangent of y=5x^2-x^3 ?
The tangent of y=5x^2-x^3 is y=(10a_{0}-3a_{0}^2)x+2a_{0}^3-5a_{0}^2