What is the tangent of x^4-32x^2+256 ?

The tangent of x^4-32x^2+256 is y=(4a_{0}^3-64a_{0})x-3a_{0}^4+32a_{0}^2+256

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tangent of x^4-32x^2+256

tangent of

x^{4} - 32 x^{2} + 256

y = (4 a_{0}^{3} - 64 a_{0}) x - 3 a_{0}^{4} + 32 a_{0}^{2} + 256

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} - 32 a_{0}^{2} + 256)

Find the slope of

y = x^{4} - 32 x^{2} + 256 : \frac{d y}{d x} = 4 x^{3} - 64 x

Compute the slope of

y = x^{4} - 32 x^{2} + 256

x = a_{0} : m = 4 a_{0}^{3} - 64 a_{0}

Find the line with slope m=

4 a_{0}^{3} - 64 a_{0}

and passing through

(a_{0}, a_{0}^{4} - 32 a_{0}^{2} + 256) : y = (4 a_{0}^{3} - 64 a_{0}) x - 3 a_{0}^{4} + 32 a_{0}^{2} + 256

y = (4 a_{0}^{3} - 64 a_{0}) x - 3 a_{0}^{4} + 32 a_{0}^{2} + 256

\frac{d}{d x} (\frac{x ^{3}}{18})

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\int \frac{4 x + 5}{4 x + 6} d x

\frac{d}{d x} (\frac{e ^{- x}}{x ^{2}})

\int 8 t^{3} (1 + t^{4})^{10} d t

What is the tangent of x^4-32x^2+256 ?
The tangent of x^4-32x^2+256 is y=(4a_{0}^3-64a_{0})x-3a_{0}^4+32a_{0}^2+256