What is the tangent of f(x)= 1/(x^3) ?

The tangent of f(x)= 1/(x^3) is y=-3/(a_{0)^4}x+4/(a_{0)^3}

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tangent of f(x)= 1/(x^3)

tangent of

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

y = - \frac{3}{a _{0}^{4}} x + \frac{4}{a _{0}^{3}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{a _{0}^{3}})

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = - \frac{3}{a _{0}^{4}}

Find the line with slope m=

- \frac{3}{a _{0}^{4}}

and passing through

(a_{0}, \frac{1}{a _{0}^{3}}) : y = - \frac{3}{a _{0}^{4}} x + \frac{4}{a _{0}^{3}}

y = - \frac{3}{a _{0}^{4}} x + \frac{4}{a _{0}^{3}}

f (x) = 20 x^{2} + 14 x - 2 x^{2}

- 12 x cos (x^{2}) + 8 x^{3} sin (x^{2})

\frac{d}{d x} (arccos (\frac{10}{x}))

\frac{x}{x + \frac{c}{x}}

f (x) = e^{x} - 6 x^{5}

What is the tangent of f(x)= 1/(x^3) ?
The tangent of f(x)= 1/(x^3) is y=-3/(a_{0)^4}x+4/(a_{0)^3}