What is the tangent of y= 2/(x^2) ?

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

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tangent of y= 2/(x^2)

tangent of

y = \frac{2}{x ^{2}}

y = - \frac{4}{a _{0}^{3}} x + \frac{6}{a _{0}^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{2}{a _{0}^{2}})

Find the slope of

y = \frac{2}{x ^{2}} : \frac{d y}{d x} = - \frac{4}{x ^{3}}

Compute the slope of

y = \frac{2}{x ^{2}}

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

Find the line with slope m=

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

and passing through

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

y = - \frac{4}{a _{0}^{3}} x + \frac{6}{a _{0}^{2}}

\int_{6}^{\infty} \frac{1}{x ^{2} + x} d x

\frac{d y}{d t} = 2 - y

\frac{d y}{d x} = x^{4} (32 - x^{5})^{5}

(- 1.6) (1.5)

ln (x), x = 1

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