Question

find the equation of the tangent line to the curve $y = \frac{5x}{1 + x^2}$ at the point $(5, \frac{25}{26})$. the equation of this tangent line can be written in the form $y = mx + b$. the equation of the tangent line is $y = -\frac{30}{169}x + \frac{625}{338}$

Explanation:

Step1: Find derivative (slope)

Use quotient rule: If $y=\frac{u}{v}$, $y'=\frac{u'v-uv'}{v^2}$.
Let $u=5x$, $u'=5$; $v=1+x^2$, $v'=2x$.

$$\begin{align*} y'&=\frac{5(1+x^2)-5x(2x)}{(1+x^2)^2}\\ &=\frac{5+5x^2-10x^2}{(1+x^2)^2}\\ &=\frac{5-5x^2}{(1+x^2)^2} \end{align*}$$

Step2: Calculate slope at $x=5$

Substitute $x=5$ into $y'$:

$$\begin{align*} y'(5)&=\frac{5-5(5)^2}{(1+(5)^2)^2}\\ &=\frac{5-125}{(1+25)^2}\\ &=\frac{-120}{26^2}\\ &=-\frac{120}{676}=-\frac{30}{169} \end{align*}$$

Step3: Solve for $b$ using point

Use point $(5, \frac{25}{26})$ and $m=-\frac{30}{169}$ in $y=mx+b$:

$$ \frac{25}{26}=-\frac{30}{169}(5)+b $$

$$ \frac{25}{26}=-\frac{150}{169}+b $$

Convert to common denominator 338:

$$ \frac{325}{338}=-\frac{300}{338}+b $$

$$ b=\frac{325+300}{338}=\frac{625}{338} $$

Answer:

$y = -\frac{30}{169}x + \frac{625}{338}$