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

The tangent of f(x)=8(x-1/x)^2,\at x=2 is y=30x-42

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

tangent of

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

y = 30 x - 42

Solution steps

Find the tangent point:

(2, 18)

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 = 2 : m = 30

Find the line with slope m=

30

and passing through

(2, 18) : y = 30 x - 42

y = 30 x - 42

\frac{d}{d x} (1 0^{- 3 x^{2}})

\int x^{2} + 49 d x

y = 2 x, y = \frac{2 x}{3}

\frac{d}{d x} (2 x + y)^{3} = x

\frac{\partial}{\partial x} (cos^{3} (x))

What is the tangent of f(x)=8(x-1/x)^2,\at x=2 ?
The tangent of f(x)=8(x-1/x)^2,\at x=2 is y=30x-42