What is the tangent of f(x)= 3/(x^2) ?

The tangent of f(x)= 3/(x^2) is y=-6/(a_{0)^3}x+9/(a_{0)^2}

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tangent of f(x)= 3/(x^2)

tangent of

f (x) = \frac{3}{x ^{2}}

y = - \frac{6}{a _{0}^{3}} x + \frac{9}{a _{0}^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{3}{a _{0}^{2}})

Find the slope of

f (x) = \frac{3}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{6}{x ^{3}}

Compute the slope of

f (x) = \frac{3}{x ^{2}}

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

Find the line with slope m=

- \frac{6}{a _{0}^{3}}

and passing through

(a_{0}, \frac{3}{a _{0}^{2}}) : y = - \frac{6}{a _{0}^{3}} x + \frac{9}{a _{0}^{2}}

y = - \frac{6}{a _{0}^{3}} x + \frac{9}{a _{0}^{2}}

\frac{\partial}{\partial x} (\frac{( a + c x )}{c + b})

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\frac{\partial}{\partial u} (u sin (v))

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

What is the tangent of f(x)= 3/(x^2) ?
The tangent of f(x)= 3/(x^2) is y=-6/(a_{0)^3}x+9/(a_{0)^2}