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

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

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tangent of y=x^3-4x

tangent of

y = 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}

\frac{2 - x}{e ^{3 x}}

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

\frac{d y}{d x} = (8 + x)^{2}

\int \frac{3 x ^{2} - x + 4}{x ^{3} + 2 x ^{2} + 6 x} d x

\frac{d}{d x} (\frac{13}{ln ( x )})

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