What is the tangent of x^3-4x ?

The tangent of x^3-4x is y=(3a_{0}^2-4)x-2a_{0}^3

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tangent of x^3-4x

tangent of

x^{3} - 4 x

y = (3 a_{0}^{2} - 4) 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}^{3} - 4 a_{0})

Find the slope of

y = x^{3} - 4 x : \frac{d y}{d x} = 3 x^{2} - 4

Compute the slope of

y = x^{3} - 4 x

x = a_{0} : m = 3 a_{0}^{2} - 4

Find the line with slope m=

3 a_{0}^{2} - 4

and passing through

(a_{0}, a_{0}^{3} - 4 a_{0}) : y = (3 a_{0}^{2} - 4) x - 2 a_{0}^{3}

y = (3 a_{0}^{2} - 4) x - 2 a_{0}^{3}

\int x (a - x)^{\frac{5}{2}} d x

\int \frac{2}{x ^{10}} d x

\frac{d}{d x} (ln (14) \cdot 4 9^{x} \cdot 2^{2 x + 1})

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

\frac{\partial}{\partial x} (\frac{e ^{x}}{e ^{x} + e ^{y}})

What is the tangent of x^3-4x ?
The tangent of x^3-4x is y=(3a_{0}^2-4)x-2a_{0}^3