What is the tangent of x^4+9x^2-x ?

The tangent of x^4+9x^2-x is y=(4a_{0}^3+18a_{0}-1)x-3a_{0}^4-9a_{0}^2

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tangent of x^4+9x^2-x

tangent of

x^{4} + 9 x^{2} - x

y = (4 a_{0}^{3} + 18 a_{0} - 1) x - 3 a_{0}^{4} - 9 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} + 9 a_{0}^{2} - a_{0})

Find the slope of

y = x^{4} + 9 x^{2} - x : \frac{d y}{d x} = 4 x^{3} + 18 x - 1

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 4 a_{0}^{3} + 18 a_{0} - 1

Find the line with slope m=

4 a_{0}^{3} + 18 a_{0} - 1

and passing through

(a_{0}, a_{0}^{4} + 9 a_{0}^{2} - a_{0}) : y = (4 a_{0}^{3} + 18 a_{0} - 1) x - 3 a_{0}^{4} - 9 a_{0}^{2}

y = (4 a_{0}^{3} + 18 a_{0} - 1) x - 3 a_{0}^{4} - 9 a_{0}^{2}

4 x^{2} y^{^{''}} + 4 x y^{^{'}} - y = 0

\int x cos^{3} (x) d x

\frac{x ^{3}}{3} + 2 x^{2} + 4 x - 1

x \to 0 lim (6 sin (x) - 1)

\frac{lo g _{10} ( 2 - x )}{x + 3}

What is the tangent of x^4+9x^2-x ?
The tangent of x^4+9x^2-x is y=(4a_{0}^3+18a_{0}-1)x-3a_{0}^4-9a_{0}^2