What is the tangent of 4/(x^5)-1/(x^2) ?

The tangent of 4/(x^5)-1/(x^2) is y=(-(20)/(a_{0)^6}+2/(a_{0)^3})x+(24)/(a_{0)^5}-3/(a_{0)^2}

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tangent of 4/(x^5)-1/(x^2)

tangent of

\frac{4}{x ^{5}} - \frac{1}{x ^{2}}

y = (- \frac{20}{a _{0}^{6}} + \frac{2}{a _{0}^{3}}) x + \frac{24}{a _{0}^{5}} - \frac{3}{a _{0}^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{4}{a _{0}^{5}} - \frac{1}{a _{0}^{2}})

Find the slope of

y = \frac{4}{x ^{5}} - \frac{1}{x ^{2}} : \frac{d y}{d x} = - \frac{20}{x ^{6}} + \frac{2}{x ^{3}}

Compute the slope of

y = \frac{4}{x ^{5}} - \frac{1}{x ^{2}}

x = a_{0} : m = - \frac{20}{a _{0}^{6}} + \frac{2}{a _{0}^{3}}

Find the line with slope m=

- \frac{20}{a _{0}^{6}} + \frac{2}{a _{0}^{3}}

and passing through

(a_{0}, \frac{4}{a _{0}^{5}} - \frac{1}{a _{0}^{2}}) : y = (- \frac{20}{a _{0}^{6}} + \frac{2}{a _{0}^{3}}) x + \frac{24}{a _{0}^{5}} - \frac{3}{a _{0}^{2}}

y = (- \frac{20}{a _{0}^{6}} + \frac{2}{a _{0}^{3}}) x + \frac{24}{a _{0}^{5}} - \frac{3}{a _{0}^{2}}

\int \frac{cos ( x )}{1 + 8 sin ( x )} d x

\frac{d y}{d x} = (\frac{x}{y})

\frac{4}{x}, 1

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

2 x^{2} - 2 x + 4

What is the tangent of 4/(x^5)-1/(x^2) ?
The tangent of 4/(x^5)-1/(x^2) is y=(-(20)/(a_{0)^6}+2/(a_{0)^3})x+(24)/(a_{0)^5}-3/(a_{0)^2}