Question

hw 10 - product and quotient rules section 2.5: problem 11 (1 point) find an equation of the tangent line to the curve y = \\(\frac{2x}{1 + x^{2}}\\) at the point (5, \\(\frac{5}{13}\\)). an equation of the tangent line is y = . round to 3 decimal places.

Explanation:

Step1: Find the derivative using quotient - rule

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

Step2: Evaluate the derivative at $x = 5$

Substitute $x = 5$ into $y^\prime$. $y^\prime(5)=\frac{2-2\times5^{2}}{(1 + 5^{2})^{2}}=\frac{2-50}{(1 + 25)^{2}}=\frac{-48}{26^{2}}=\frac{-48}{676}=-\frac{12}{169}\approx - 0.071$.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(5,\frac{5}{13})$ and $m =-\frac{12}{169}$. So $y-\frac{5}{13}=-\frac{12}{169}(x - 5)$.

Step4: Simplify the equation

$y-\frac{5}{13}=-\frac{12}{169}x+\frac{60}{169}$. Multiply through by 169 to clear the fractions: $169y-65=-12x + 60$. Then $169y=-12x+125$, and $y=-\frac{12}{169}x+\frac{125}{169}\approx - 0.071x + 0.739$.

Answer:

$y=-0.071x + 0.739$