What is the tangent of y=x^3-9x,(1,-8) ?

The tangent of y=x^3-9x,(1,-8) is y=-6x-2

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tangent of y=x^3-9x,(1,-8)

tangent of

y = x^{3} - 9 x, (1, - 8)

y = - 6 x - 2

Solution steps

Find the slope of

y = x^{3} - 9 x : \frac{d y}{d x} = 3 x^{2} - 9

Compute the slope of

y = x^{3} - 9 x

(1, - 8) : m = - 6

Find the line with slope m=

- 6

and passing through

(1, - 8) : y = - 6 x - 2

y = - 6 x - 2

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

\frac{\partial}{\partial x} (e^{s i n (x^{2})})

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

x^{3} y^{^{'}} + 2 x^{2} y = - 3 cos (3 x) y^{- \frac{1}{2}}

y^{^{'}} + 2 x y - 6 = 0

What is the tangent of y=x^3-9x,(1,-8) ?
The tangent of y=x^3-9x,(1,-8) is y=-6x-2