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

The tangent of y=x^4+4x^2-x is y=(4a_{0}^3+8a_{0}-1)x-3a_{0}^4-4a_{0}^2

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

tangent of

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

y = (4 a_{0}^{3} + 8 a_{0} - 1) x - 3 a_{0}^{4} - 4 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} + 4 a_{0}^{2} - a_{0})

Find the slope of

y = x^{4} + 4 x^{2} - x : \frac{d y}{d x} = 4 x^{3} + 8 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} + 8 a_{0} - 1

Find the line with slope m=

4 a_{0}^{3} + 8 a_{0} - 1

and passing through

(a_{0}, a_{0}^{4} + 4 a_{0}^{2} - a_{0}) : y = (4 a_{0}^{3} + 8 a_{0} - 1) x - 3 a_{0}^{4} - 4 a_{0}^{2}

y = (4 a_{0}^{3} + 8 a_{0} - 1) x - 3 a_{0}^{4} - 4 a_{0}^{2}

\frac{d}{d x} (ln (x - 8))

\int \frac{cos ( x )}{sin ^{2} ( x ) - 9} d x

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

\frac{d}{d x} (lo g_{3} (2 x - 1))

\frac{d}{d t} (\frac{1}{t ^{3}})

What is the tangent of y=x^4+4x^2-x ?
The tangent of y=x^4+4x^2-x is y=(4a_{0}^3+8a_{0}-1)x-3a_{0}^4-4a_{0}^2