What is the tangent of y= 8/(x^2) ?

The tangent of y= 8/(x^2) is y=-(16)/(a_{0)^3}x+(24)/(a_{0)^2}

Popular >

tangent of y= 8/(x^2)

tangent of

y = \frac{8}{x ^{2}}

y = - \frac{16}{a _{0}^{3}} x + \frac{24}{a _{0}^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{8}{a _{0}^{2}})

Find the slope of

y = \frac{8}{x ^{2}} : \frac{d y}{d x} = - \frac{16}{x ^{3}}

Compute the slope of

y = \frac{8}{x ^{2}}

x = a_{0} : m = - \frac{16}{a _{0}^{3}}

Find the line with slope m=

- \frac{16}{a _{0}^{3}}

and passing through

(a_{0}, \frac{8}{a _{0}^{2}}) : y = - \frac{16}{a _{0}^{3}} x + \frac{24}{a _{0}^{2}}

y = - \frac{16}{a _{0}^{3}} x + \frac{24}{a _{0}^{2}}

\frac{d}{d x} (16 e^{2 x})

y = 3 arccos (\frac{x}{2})

\frac{d y}{d x} = \frac{4 sec ( y )}{( x + 5 ) ^{2}}

(- 2.8) (- 22.8)

\frac{d}{d x} (56 x^{6})

What is the tangent of y= 8/(x^2) ?
The tangent of y= 8/(x^2) is y=-(16)/(a_{0)^3}x+(24)/(a_{0)^2}