What is the tangent of y=x^4+6x^2-x ?

The tangent of y=x^4+6x^2-x is y=(4a_{0}^3+12a_{0}-1)x-3a_{0}^4-6a_{0}^2

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tangent of y=x^4+6x^2-x

tangent of

y = x^{4} + 6 x^{2} - x

y = (4 a_{0}^{3} + 12 a_{0} - 1) x - 3 a_{0}^{4} - 6 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} + 6 a_{0}^{2} - a_{0})

Find the slope of

y = x^{4} + 6 x^{2} - x : \frac{d y}{d x} = 4 x^{3} + 12 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} + 12 a_{0} - 1

Find the line with slope m=

4 a_{0}^{3} + 12 a_{0} - 1

and passing through

(a_{0}, a_{0}^{4} + 6 a_{0}^{2} - a_{0}) : y = (4 a_{0}^{3} + 12 a_{0} - 1) x - 3 a_{0}^{4} - 6 a_{0}^{2}

y = (4 a_{0}^{3} + 12 a_{0} - 1) x - 3 a_{0}^{4} - 6 a_{0}^{2}

\int (4 x^{3} + x^{2} + 5) d x

\frac{\partial}{\partial y} (a x^{2} + b y^{2})

\int x (1 + x)^{\frac{2}{3}} d x

y = cos (x), x = \frac{π}{2}, x = 2 π, y = 0

n = 0 \sum \infty \frac{2}{3} (\frac{1}{3})^{n}

What is the tangent of y=x^4+6x^2-x ?
The tangent of y=x^4+6x^2-x is y=(4a_{0}^3+12a_{0}-1)x-3a_{0}^4-6a_{0}^2