What is the tangent of y=x^4+9x^2-x(1.9) ?

The tangent of y=x^4+9x^2-x(1.9) is y=(4a_{0}^3+18a_{0}-1.9)x-3a_{0}^4-9a_{0}^2

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tangent of y=x^4+9x^2-x(1.9)

tangent of

y = x^{4} + 9 x^{2} - x (1.9)

y = (4 a_{0}^{3} + 18 a_{0} - 1.9) x - 3 a_{0}^{4} - 9 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} + 9 a_{0}^{2} - a_{0} \cdot 1.9)

Find the slope of

y = x^{4} + 9 x^{2} - x 1.9 : \frac{d y}{d x} = 4 x^{3} + 18 x - 1.9

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} + 18 a_{0} - 1.9

Find the line with slope m=

4 a_{0}^{3} + 18 a_{0} - 1.9

and passing through

(a_{0}, a_{0}^{4} + 9 a_{0}^{2} - a_{0} 1.9) : y = (4 a_{0}^{3} + 18 a_{0} - 1.9) x - 3 a_{0}^{4} - 9 a_{0}^{2}

y = (4 a_{0}^{3} + 18 a_{0} - 1.9) x - 3 a_{0}^{4} - 9 a_{0}^{2}

2 cos (x) + 3 sin (x), at x = \frac{π}{2}

\frac{ln ( 2 x )}{2 x}

y = 20 x, y = x^{2}

(\frac{1}{2})^{^{'}}

\frac{\partial}{\partial x} (\frac{sin ( x )}{cos ( x )})

What is the tangent of y=x^4+9x^2-x(1.9) ?
The tangent of y=x^4+9x^2-x(1.9) is y=(4a_{0}^3+18a_{0}-1.9)x-3a_{0}^4-9a_{0}^2