Question

current objective find the derivative of other trigonometric functions question find the derivative of $f(x)=-2csc(x)+4cot(x)$. provide your answer below: $f(x)=square$

Explanation:

Step1: Recall derivative rules

The derivative of $\csc(x)$ is $-\csc(x)\cot(x)$ and the derivative of $\cot(x)$ is $-\csc^{2}(x)$. Also, use the sum - rule of derivatives $(u + v)'=u'+v'$ and the constant - multiple rule $(cf(x))' = cf'(x)$ where $c$ is a constant.

Step2: Differentiate $- 2\csc(x)$

Using the constant - multiple rule and the derivative of $\csc(x)$, we have $(-2\csc(x))'=-2(-\csc(x)\cot(x)) = 2\csc(x)\cot(x)$.

Step3: Differentiate $4\cot(x)$

Using the constant - multiple rule and the derivative of $\cot(x)$, we get $(4\cot(x))'=4(-\csc^{2}(x))=-4\csc^{2}(x)$.

Step4: Apply sum - rule

$f'(x)=(-2\csc(x))'+(4\cot(x))'$. Substituting the results from Step 2 and Step 3, we have $f'(x)=2\csc(x)\cot(x)-4\csc^{2}(x)$.

Answer:

$2\csc(x)\cot(x)-4\csc^{2}(x)$