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

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

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

tangent of

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

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

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = 4 a_{0}^{3} + 4 a_{0} - 1

Find the line with slope m=

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

and passing through

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

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

\frac{d}{d x} (\frac{1}{5 x ^{4}} - \frac{5}{2 x})

f (x) = 6 x^{2}, (2, 24)

\int 1 - 4 y d y

\int_{0}^{1} ln (8 x) d x

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

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