What is the tangent of y= 5/x ?

The tangent of y= 5/x is y=-5/(a_{0)^2}x+(10)/(a_{0)}

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tangent of y= 5/x

tangent of

y = \frac{5}{x}

y = - \frac{5}{a _{0}^{2}} x + \frac{10}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{5}{a _{0}})

Find the slope of

y = \frac{5}{x} : \frac{d y}{d x} = - \frac{5}{x ^{2}}

Compute the slope of

y = \frac{5}{x}

x = a_{0} : m = - \frac{5}{a _{0}^{2}}

Find the line with slope m=

- \frac{5}{a _{0}^{2}}

and passing through

(a_{0}, \frac{5}{a _{0}}) : y = - \frac{5}{a _{0}^{2}} x + \frac{10}{a _{0}}

y = - \frac{5}{a _{0}^{2}} x + \frac{10}{a _{0}}

y^{^{'}} + sin (x) y = - sin (x)

\int x \cdot sin (x) d x

f (x) = - 0.5 x^{2}, at x = 10

\int_{0}^{1} 4 (1 + x)^{5} d x

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

What is the tangent of y= 5/x ?
The tangent of y= 5/x is y=-5/(a_{0)^2}x+(10)/(a_{0)}