Question

find the derivative of $\frac{1}{2}csc 2x$

Explanation:

Step1: Recall the constant - multiple rule

The constant - multiple rule states that if $y = cf(x)$, then $y'=cf'(x)$. Here $c=\frac{1}{2}$ and $f(x)=\csc(2x)$. So $y'=\frac{1}{2}(\csc(2x))'$.

Step2: Apply the chain - rule to find $(\csc(2x))'$

The derivative of $\csc(u)$ with respect to $u$ is $-\csc(u)\cot(u)$. Let $u = 2x$. Then by the chain - rule $\frac{d}{dx}\csc(u)=-\csc(u)\cot(u)\cdot\frac{du}{dx}$. Since $\frac{du}{dx}=2$, we have $(\csc(2x))'=-\csc(2x)\cot(2x)\cdot2=- 2\csc(2x)\cot(2x)$.

Step3: Calculate $y'$

Substitute $(\csc(2x))'$ into the expression from Step1. $y'=\frac{1}{2}\times(-2\csc(2x)\cot(2x))=-\csc(2x)\cot(2x)$.

Answer:

$-\csc(2x)\cot(2x)$