What is the tangent of-2x^4+5x^2-3(1) ?

The tangent of-2x^4+5x^2-3(1) is y=(-8a_{0}^3+10a_{0})x+6a_{0}^4-5a_{0}^2-3

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tangent of-2x^4+5x^2-3(1)

tangent of

- 2 x^{4} + 5 x^{2} - 3 (1)

y = (- 8 a_{0}^{3} + 10 a_{0}) x + 6 a_{0}^{4} - 5 a_{0}^{2} - 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 2 a_{0}^{4} + 5 a_{0}^{2} - 3)

Find the slope of

y = - 2 x^{4} + 5 x^{2} - 3 \cdot 1 : \frac{d y}{d x} = - 8 x^{3} + 10 x

Compute the slope of

y = - 2 x^{4} + 5 x^{2} - 3 \cdot 1

x = a_{0} : m = - 8 a_{0}^{3} + 10 a_{0}

Find the line with slope m=

- 8 a_{0}^{3} + 10 a_{0}

and passing through

(a_{0}, - 2 a_{0}^{4} + 5 a_{0}^{2} - 3) : y = (- 8 a_{0}^{3} + 10 a_{0}) x + 6 a_{0}^{4} - 5 a_{0}^{2} - 3

y = (- 8 a_{0}^{3} + 10 a_{0}) x + 6 a_{0}^{4} - 5 a_{0}^{2} - 3

y = e^{t}

x^{2} \frac{d y}{d x} = y

f (x) = 12 x, at x = 9

(t^{2} + 4) y^{^{'}} + 2 t y = t^{2} (t^{2} + 4)

y = x^{2} + 3, (4, 19)

What is the tangent of-2x^4+5x^2-3(1) ?
The tangent of-2x^4+5x^2-3(1) is y=(-8a_{0}^3+10a_{0})x+6a_{0}^4-5a_{0}^2-3