What is the tangent of x^4-2x^2+3 ?

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

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tangent of x^4-2x^2+3

tangent of

x^{4} - 2 x^{2} + 3

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

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} + 3)

Find the slope of

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

Compute the slope of

y = x^{4} - 2 x^{2} + 3

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

Find the line with slope m=

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

and passing through

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

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

x \to 7 lim (- 7)

\frac{\partial}{\partial z} (e^{7 yz})

\frac{d}{d x} (x^{3} - 3 x^{2} + 4 x - 1)

y^{^{'}} + sin (t) y = sin (t)

\frac{d}{d x} (\frac{1}{2} x^{\frac{3}{2}})

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