What is the tangent of (4x)/(x+1),\at (3.3) ?

The tangent of (4x)/(x+1),\at (3.3) is y= 4/((a_{0)+1)^2}x+(4a_{0}^2)/((a_{0)+1)^2}

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tangent of (4x)/(x+1),\at (3.3)

tangent of

\frac{4 x}{x + 1}, at (3.3)

y = \frac{4}{( a _{0} + 1 ) ^{2}} x + \frac{4 a _{0}^{2}}{( a _{0} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{4 a _{0}}{a _{0} + 1})

Find the slope of

y = \frac{4 x}{x + 1} : \frac{d y}{d x} = \frac{4}{( x + 1 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{4}{( a _{0} + 1 ) ^{2}}

Find the line with slope m=

\frac{4}{( a _{0} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{4 a _{0}}{a _{0} + 1}) : y = \frac{4}{( a _{0} + 1 ) ^{2}} x + \frac{4 a _{0}^{2}}{( a _{0} + 1 ) ^{2}}

y = \frac{4}{( a _{0} + 1 ) ^{2}} x + \frac{4 a _{0}^{2}}{( a _{0} + 1 ) ^{2}}

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

x \to - 4 lim (\frac{5}{x + 4})

\frac{d u}{1 + sin ( u )} = d x

5 t \cdot \frac{d y}{d t} - 5 y = t

f (x) = x^{x^{2}}

What is the tangent of (4x)/(x+1),\at (3.3) ?
The tangent of (4x)/(x+1),\at (3.3) is y= 4/((a_{0)+1)^2}x+(4a_{0}^2)/((a_{0)+1)^2}