What is the tangent of f(x)=8(x-1/x)^2 ?

The tangent of f(x)=8(x-1/x)^2 is y=8(2a_{0}-2/(a_{0)^3})x-8a_{0}^2+(24)/(a_{0)^2}-16

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tangent of f(x)=8(x-1/x)^2

tangent of

f (x) = 8 (x - \frac{1}{x})^{2}

y = 8 (2 a_{0} - \frac{2}{a _{0}^{3}}) x - 8 a_{0}^{2} + \frac{24}{a _{0}^{2}} - 16

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 (a_{0}^{2} - 2 + \frac{1}{a _{0}^{2}}))

Find the slope of

f (x) = 8 (x - \frac{1}{x})^{2} : \frac{df ( x )}{d x} = 8 (2 x - \frac{2}{x ^{3}})

Compute the slope of

f (x) = 8 (x - \frac{1}{x})^{2}

x = a_{0} : m = 8 (2 a_{0} - \frac{2}{a _{0}^{3}})

Find the line with slope m=

8 (2 a_{0} - \frac{2}{a _{0}^{3}})

and passing through

(a_{0}, 8 (a_{0}^{2} - 2 + \frac{1}{a _{0}^{2}})) : y = 8 (2 a_{0} - \frac{2}{a _{0}^{3}}) x - 8 a_{0}^{2} + \frac{24}{a _{0}^{2}} - 16

y = 8 (2 a_{0} - \frac{2}{a _{0}^{3}}) x - 8 a_{0}^{2} + \frac{24}{a _{0}^{2}} - 16

\int \frac{3}{4} d x

\int_{2}^{4} 2 π x (5 - x) d x

\int sin (x^{3}) d x

\frac{d}{d x} (6 sin (6 x))

- 3

What is the tangent of f(x)=8(x-1/x)^2 ?
The tangent of f(x)=8(x-1/x)^2 is y=8(2a_{0}-2/(a_{0)^3})x-8a_{0}^2+(24)/(a_{0)^2}-16