find $\frac{dy}{d\theta}$ for $y = \frac{2\theta}{\theta^{2}-5}$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{d\theta}=\frac{u'v - uv'}{v^{2}}$. Here, $u = 2\theta$, so $u'=2$, and $v=\theta^{2}-5$, so $v' = 2\theta$.

Step2: Substitute values into quotient - rule

$\frac{dy}{d\theta}=\frac{2(\theta^{2}-5)-2\theta\times(2\theta)}{(\theta^{2}-5)^{2}}$.

Step3: Simplify the numerator

$2(\theta^{2}-5)-4\theta^{2}=2\theta^{2}-10 - 4\theta^{2}=-2\theta^{2}-10$.

Answer:

$\frac{-2\theta^{2}-10}{(\theta^{2}-5)^{2}}$