What is the tangent of y=x^4+5x^2-x ?

The tangent of y=x^4+5x^2-x is y=(4a_{0}^3+10a_{0}-1)x-3a_{0}^4-5a_{0}^2

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tangent of y=x^4+5x^2-x

tangent of

y = x^{4} + 5 x^{2} - x

y = (4 a_{0}^{3} + 10 a_{0} - 1) x - 3 a_{0}^{4} - 5 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} + 5 a_{0}^{2} - a_{0})

Find the slope of

y = x^{4} + 5 x^{2} - x : \frac{d y}{d x} = 4 x^{3} + 10 x - 1

Compute the slope of

y = x^{4} + 5 x^{2} - x

x = a_{0} : m = 4 a_{0}^{3} + 10 a_{0} - 1

Find the line with slope m=

4 a_{0}^{3} + 10 a_{0} - 1

and passing through

(a_{0}, a_{0}^{4} + 5 a_{0}^{2} - a_{0}) : y = (4 a_{0}^{3} + 10 a_{0} - 1) x - 3 a_{0}^{4} - 5 a_{0}^{2}

y = (4 a_{0}^{3} + 10 a_{0} - 1) x - 3 a_{0}^{4} - 5 a_{0}^{2}

\int cos (\frac{nπ x}{2}) d x

x \to 3 + lim (\frac{3}{x - 3} - 2 x)

\frac{d}{d x} (ln (3 x + 5 y))

\frac{d}{d x} (\frac{1}{4 x - 7})

x \to 0 + lim ((1 + tan (x))^{\frac{3}{x}})

What is the tangent of y=x^4+5x^2-x ?
The tangent of y=x^4+5x^2-x is y=(4a_{0}^3+10a_{0}-1)x-3a_{0}^4-5a_{0}^2