What is the tangent of f(x)=8(x-1/x)^3,\at x=2 ?

The tangent of f(x)=8(x-1/x)^3,\at x=2 is y= 135/2 x-108

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tangent of f(x)=8(x-1/x)^3,\at x=2

tangent of

f (x) = 8 (x - \frac{1}{x})^{3}, at x = 2

y = \frac{135}{2} x - 108

Solution steps

Find the tangent point:

(2, 27)

Find the slope of

f (x) = 8 (x - \frac{1}{x})^{3} : \frac{df ( x )}{d x} = 24 x^{2} + \frac{24}{x ^{4}} - \frac{24}{x ^{2}} - 24

Compute the slope of

f (x) = 8 (x - \frac{1}{x})^{3}

x = 2 : m = \frac{135}{2}

Find the line with slope m=

\frac{135}{2}

and passing through

(2, 27) : y = \frac{135}{2} x - 108

y = \frac{135}{2} x - 108

n = 0 \sum \infty (3 x)^{n}

y = cot (x), at x = \frac{π}{4}

\frac{d}{d x} (sin (a x^{2}))

\int t^{3} + 1 d t

\int_{6}^{8} \frac{88}{( x - 6 ) ^{3}} d x

What is the tangent of f(x)=8(x-1/x)^3,\at x=2 ?
The tangent of f(x)=8(x-1/x)^3,\at x=2 is y= 135/2 x-108