What is the tangent of y=x^4+2x^2-x(1.2) ?

The tangent of y=x^4+2x^2-x(1.2) is y=(4a_{0}^3+4a_{0}-1.2)x-3a_{0}^4-2a_{0}^2

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tangent of y=x^4+2x^2-x(1.2)

tangent of

y = x^{4} + 2 x^{2} - x (1.2)

y = (4 a_{0}^{3} + 4 a_{0} - 1.2) 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} \cdot 1.2)

Find the slope of

y = x^{4} + 2 x^{2} - x 1.2 : \frac{d y}{d x} = 4 x^{3} + 4 x - 1.2

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} + 4 a_{0} - 1.2

Find the line with slope m=

4 a_{0}^{3} + 4 a_{0} - 1.2

and passing through

(a_{0}, a_{0}^{4} + 2 a_{0}^{2} - a_{0} 1.2) : y = (4 a_{0}^{3} + 4 a_{0} - 1.2) x - 3 a_{0}^{4} - 2 a_{0}^{2}

y = (4 a_{0}^{3} + 4 a_{0} - 1.2) x - 3 a_{0}^{4} - 2 a_{0}^{2}

(x^{3} - 25 x)^{12}

\frac{\partial}{\partial x} (ln (2 x + 4 y + 9 z))

f (x) = \frac{( 8 x )}{( sin ( x ) + cos ( x ) )}

\frac{d}{d x} ((4 x)^{2 x})

y = (6 x^{2} - 6 x + 3) (1 + 2 x), at x = 1

What is the tangent of y=x^4+2x^2-x(1.2) ?
The tangent of y=x^4+2x^2-x(1.2) is y=(4a_{0}^3+4a_{0}-1.2)x-3a_{0}^4-2a_{0}^2