find an equation of the tangent line to the given curve at the specified point. y = 4x / (x + 1), (3, 3) y =

Explanation:

Step1: Find the derivative using quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 4x$, $u'=4$, $v=x + 1$, $v'=1$. So $y'=\frac{4(x + 1)-4x\times1}{(x + 1)^{2}}=\frac{4x+4 - 4x}{(x + 1)^{2}}=\frac{4}{(x + 1)^{2}}$.

Step2: Evaluate the derivative at the given $x$ - value

We want to find the slope of the tangent line at $x = 3$. Substitute $x = 3$ into $y'$. So $m=y'(3)=\frac{4}{(3 + 1)^{2}}=\frac{4}{16}=\frac{1}{4}$.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(3,3)$ and $m=\frac{1}{4}$. So $y - 3=\frac{1}{4}(x - 3)$.

Step4: Rearrange the equation to slope - intercept form

$y-3=\frac{1}{4}x-\frac{3}{4}$. Then $y=\frac{1}{4}x-\frac{3}{4}+3=\frac{1}{4}x+\frac{9}{4}$.

Answer:

$y=\frac{1}{4}x+\frac{9}{4}$