What is the tangent of y= 8/(x+1) ?

The tangent of y= 8/(x+1) is y=-8/((a_{0)+1)^2}x+(16a_{0}+8)/((a_{0)+1)^2}

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tangent of y= 8/(x+1)

tangent of

y = \frac{8}{x + 1}

y = - \frac{8}{( a _{0} + 1 ) ^{2}} x + \frac{16 a _{0} + 8}{( a _{0} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{8}{a _{0} + 1})

Find the slope of

y = \frac{8}{x + 1} : \frac{d y}{d x} = - \frac{8}{( x + 1 ) ^{2}}

Compute the slope of

y = \frac{8}{x + 1}

x = a_{0} : m = - \frac{8}{( a _{0} + 1 ) ^{2}}

Find the line with slope m=

- \frac{8}{( a _{0} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{8}{a _{0} + 1}) : y = - \frac{8}{( a _{0} + 1 ) ^{2}} x + \frac{16 a _{0} + 8}{( a _{0} + 1 ) ^{2}}

y = - \frac{8}{( a _{0} + 1 ) ^{2}} x + \frac{16 a _{0} + 8}{( a _{0} + 1 ) ^{2}}

x \to 0 lim (x \cdot e^{\frac{1}{x}})

y = (x^{4} - 6 x) e^{x}

\frac{\partial}{\partial x} (\frac{2 x}{x ^{2} + y ^{2} + 16})

\frac{d}{d x} (\frac{cos ( 2 x )}{x})

3 x^{5} x + \frac{7}{x ^{3} x}

What is the tangent of y= 8/(x+1) ?
The tangent of y= 8/(x+1) is y=-8/((a_{0)+1)^2}x+(16a_{0}+8)/((a_{0)+1)^2}