What is the tangent of y=4(x-1/x)^3,\at x=2 ?

The tangent of y=4(x-1/x)^3,\at x=2 is y= 135/4 x-54

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tangent of y=4(x-1/x)^3,\at x=2

tangent of

y = 4 (x - \frac{1}{x})^{3}, at x = 2

y = \frac{135}{4} x - 54

Solution steps

Find the tangent point:

(2, \frac{27}{2})

Find the slope of

y = 4 (x - \frac{1}{x})^{3} : \frac{d y}{d x} = 12 x^{2} + \frac{12}{x ^{4}} - \frac{12}{x ^{2}} - 12

Compute the slope of

y = 4 (x - \frac{1}{x})^{3}

x = 2 : m = \frac{135}{4}

Find the line with slope m=

\frac{135}{4}

and passing through

(2, \frac{27}{2}) : y = \frac{135}{4} x - 54

y = \frac{135}{4} x - 54

\int \frac{1}{( x - 1 ) 4 x ^{2} - 8 x + 3} d x

\int x + \frac{2}{sin ^{2} ( x )} d x

y^{^{''}} - y = 0, y (0) = 0, y^{^{'}} (0) = 1

\frac{d}{d x} (8 cos^{4} (x))

x \to - 1 lim (f (x))

What is the tangent of y=4(x-1/x)^3,\at x=2 ?
The tangent of y=4(x-1/x)^3,\at x=2 is y= 135/4 x-54