What is the tangent of x^3-7x ?

The tangent of x^3-7x is y=(3a_{0}^2-7)x-2a_{0}^3

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tangent of x^3-7x

tangent of

x^{3} - 7 x

y = (3 a_{0}^{2} - 7) 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} - 7 a_{0})

Find the slope of

y = x^{3} - 7 x : \frac{d y}{d x} = 3 x^{2} - 7

Compute the slope of

y = x^{3} - 7 x

x = a_{0} : m = 3 a_{0}^{2} - 7

Find the line with slope m=

3 a_{0}^{2} - 7

and passing through

(a_{0}, a_{0}^{3} - 7 a_{0}) : y = (3 a_{0}^{2} - 7) x - 2 a_{0}^{3}

y = (3 a_{0}^{2} - 7) x - 2 a_{0}^{3}

\int sin (3 x y) d x

f (x) = x + x

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

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

(\frac{d x}{d y}) = k (a - x) (a - x)

What is the tangent of x^3-7x ?
The tangent of x^3-7x is y=(3a_{0}^2-7)x-2a_{0}^3