Question

find the equation of the tangent line to the curve $y = \frac{5x}{1 + x^{2}}$ at $x = 4.$
$y=$

Explanation:

Step1: Find the point on the curve

When $x = 4$, $y=\frac{5\times4}{1 + 4^{2}}=\frac{20}{1 + 16}=\frac{20}{17}$. So the point $(x_0,y_0)=(4,\frac{20}{17})$.

Step2: Differentiate the function using the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 5x$, $u^\prime=5$, $v = 1 + x^{2}$, $v^\prime = 2x$. So $y^\prime=\frac{5(1 + x^{2})-5x\times(2x)}{(1 + x^{2})^{2}}=\frac{5 + 5x^{2}-10x^{2}}{(1 + x^{2})^{2}}=\frac{5 - 5x^{2}}{(1 + x^{2})^{2}}$.

Step3: Find the slope of the tangent line at $x = 4$

Substitute $x = 4$ into $y^\prime$: $y^\prime|_{x = 4}=\frac{5-5\times4^{2}}{(1 + 4^{2})^{2}}=\frac{5-80}{(1 + 16)^{2}}=\frac{-75}{289}$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_0=m(x - x_0)$. Substitute $m =-\frac{75}{289}$, $x_0 = 4$, and $y_0=\frac{20}{17}$ into it:
$y-\frac{20}{17}=-\frac{75}{289}(x - 4)$.
$y-\frac{20}{17}=-\frac{75}{289}x+\frac{300}{289}$.
$y=-\frac{75}{289}x+\frac{300}{289}+\frac{20}{17}$.
$y=-\frac{75}{289}x+\frac{300+340}{289}$.
$y=-\frac{75}{289}x+\frac{640}{289}$.

Answer:

$y =-\frac{75}{289}x+\frac{640}{289}$