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

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

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tangent of y=x-x^3

tangent of

y = x - x^{3}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = x - x^{3}

x = a_{0} : m = 1 - 3 a_{0}^{2}

Find the line with slope m=

1 - 3 a_{0}^{2}

and passing through

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

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

\int (cos (x))^{5} d x

\frac{\partial}{\partial y} (\frac{x - y}{x + z})

\frac{d y}{d x} = 7 x y

x \to 0 lim (3 x + 4)

\int \frac{cos ^{3} ( x ) + 4}{cos ^{2} ( x )} d x

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