What is the tangent of f(x)= 3/(x+1) ?

The tangent of f(x)= 3/(x+1) is y=-3/((a_{0)+1)^2}x+(6a_{0}+3)/((a_{0)+1)^2}

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tangent of f(x)= 3/(x+1)

tangent of

f (x) = \frac{3}{x + 1}

y = - \frac{3}{( a _{0} + 1 ) ^{2}} x + \frac{6 a _{0} + 3}{( a _{0} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{3}{a _{0} + 1})

Find the slope of

f (x) = \frac{3}{x + 1} : \frac{df ( x )}{d x} = - \frac{3}{( x + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{3}{x + 1}

x = a_{0} : m = - \frac{3}{( a _{0} + 1 ) ^{2}}

Find the line with slope m=

- \frac{3}{( a _{0} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{3}{a _{0} + 1}) : y = - \frac{3}{( a _{0} + 1 ) ^{2}} x + \frac{6 a _{0} + 3}{( a _{0} + 1 ) ^{2}}

y = - \frac{3}{( a _{0} + 1 ) ^{2}} x + \frac{6 a _{0} + 3}{( a _{0} + 1 ) ^{2}}

\frac{\partial}{\partial z} (x y cos (x + yz))

\frac{d y}{d x} = 20 - 5 y

\int \frac{2 - 8 x}{1 - 4 x ^{2}} d x

x \to 1 lim (3 x^{3} + 2 x^{2} + 4)

\int \frac{x ^{5}}{1 - x ^{2}} d x

What is the tangent of f(x)= 3/(x+1) ?
The tangent of f(x)= 3/(x+1) is y=-3/((a_{0)+1)^2}x+(6a_{0}+3)/((a_{0)+1)^2}