What is the tangent of x^4-8x^2+16 ?

The tangent of x^4-8x^2+16 is y=(4a_{0}^3-16a_{0})x-3a_{0}^4+8a_{0}^2+16

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tangent of x^4-8x^2+16

tangent of

x^{4} - 8 x^{2} + 16

y = (4 a_{0}^{3} - 16 a_{0}) x - 3 a_{0}^{4} + 8 a_{0}^{2} + 16

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} - 8 a_{0}^{2} + 16)

Find the slope of

y = x^{4} - 8 x^{2} + 16 : \frac{d y}{d x} = 4 x^{3} - 16 x

Compute the slope of

y = x^{4} - 8 x^{2} + 16

x = a_{0} : m = 4 a_{0}^{3} - 16 a_{0}

Find the line with slope m=

4 a_{0}^{3} - 16 a_{0}

and passing through

(a_{0}, a_{0}^{4} - 8 a_{0}^{2} + 16) : y = (4 a_{0}^{3} - 16 a_{0}) x - 3 a_{0}^{4} + 8 a_{0}^{2} + 16

y = (4 a_{0}^{3} - 16 a_{0}) x - 3 a_{0}^{4} + 8 a_{0}^{2} + 16

x^{2} + 2 x

\int \frac{1}{9 e ^{- 2 x} - 1} d x

f (x) = x^{3}, at x = - 1

\frac{d}{d x} (\frac{24}{( 1 + x ) ^{5}})

y^{'} = y^{2} - 2 y

What is the tangent of x^4-8x^2+16 ?
The tangent of x^4-8x^2+16 is y=(4a_{0}^3-16a_{0})x-3a_{0}^4+8a_{0}^2+16