Question

1. using the quotient rule, show that $\frac{d}{d\theta}cot(\theta)=-csc^{2}(\theta)$.

Explanation:

Step1: Recall cotangent definition

$\cot(\theta)=\frac{\cos(\theta)}{\sin(\theta)}$

Step2: Recall quotient - rule

The quotient - rule states that if $y = \frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = \cos(\theta)$ and $v=\sin(\theta)$. So $u^\prime=-\sin(\theta)$ and $v^\prime=\cos(\theta)$.

Step3: Apply quotient - rule

$\frac{d}{d\theta}\cot(\theta)=\frac{-\sin(\theta)\cdot\sin(\theta)-\cos(\theta)\cdot\cos(\theta)}{\sin^{2}(\theta)}$

Step4: Simplify the numerator

$-\sin^{2}(\theta)-\cos^{2}(\theta)=-(\sin^{2}(\theta)+\cos^{2}(\theta))$. By the Pythagorean identity $\sin^{2}(\theta)+\cos^{2}(\theta) = 1$, the numerator is $- 1$.

Step5: Simplify the expression

$\frac{-1}{\sin^{2}(\theta)}=-\csc^{2}(\theta)$

Answer:

We have shown that $\frac{d}{d\theta}\cot(\theta)=-\csc^{2}(\theta)$ using the quotient - rule.