What is the tangent of x^4+8x^2-x ?

The tangent of x^4+8x^2-x is y=(4a_{0}^3+16a_{0}-1)x-3a_{0}^4-8a_{0}^2

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tangent of x^4+8x^2-x

tangent of

x^{4} + 8 x^{2} - x

y = (4 a_{0}^{3} + 16 a_{0} - 1) x - 3 a_{0}^{4} - 8 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} + 8 a_{0}^{2} - a_{0})

Find the slope of

y = x^{4} + 8 x^{2} - x : \frac{d y}{d x} = 4 x^{3} + 16 x - 1

Compute the slope of

y = x^{4} + 8 x^{2} - x

x = a_{0} : m = 4 a_{0}^{3} + 16 a_{0} - 1

Find the line with slope m=

4 a_{0}^{3} + 16 a_{0} - 1

and passing through

(a_{0}, a_{0}^{4} + 8 a_{0}^{2} - a_{0}) : y = (4 a_{0}^{3} + 16 a_{0} - 1) x - 3 a_{0}^{4} - 8 a_{0}^{2}

y = (4 a_{0}^{3} + 16 a_{0} - 1) x - 3 a_{0}^{4} - 8 a_{0}^{2}

f^{^{'}} (x) = x^{3} + 6 x^{2}

\frac{d y}{d x} = y^{2} x, y (0) = 6

y = 8 ln (x), y = x ln (x)

(\frac{2}{π}) sin (π t), t

\frac{d}{d x} ((sin^{2} (x))^{2})

What is the tangent of x^4+8x^2-x ?
The tangent of x^4+8x^2-x is y=(4a_{0}^3+16a_{0}-1)x-3a_{0}^4-8a_{0}^2