Question

question
find the derivative of $f(x)=-2cot(x)+6x^{4}csc(x)$.
provide your answer below:
$f(x)= square$

Explanation:

Step1: Recall derivative rules

The derivative of $\cot(x)$ is $-\csc^{2}(x)$ and the derivative of a sum $u + v$ is $u'+v'$. Also, use the product - rule $(uv)'=u'v + uv'$ for $u = 6x^{4}$ and $v=\csc(x)$.

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

The derivative of $-2\cot(x)$ is $-2\times(-\csc^{2}(x)) = 2\csc^{2}(x)$ since the derivative of $\cot(x)$ is $-\csc^{2}(x)$ and by the constant - multiple rule $(cf(x))'=cf'(x)$.

Step3: Differentiate $6x^{4}\csc(x)$ using product - rule

The derivative of $u = 6x^{4}$ is $u'=24x^{3}$ (by the power rule $(x^{n})'=nx^{n - 1}$) and the derivative of $v=\csc(x)$ is $v'=-\csc(x)\cot(x)$. Then $(uv)'=u'v+uv'=24x^{3}\csc(x)+6x^{4}(-\csc(x)\cot(x))=24x^{3}\csc(x)-6x^{4}\csc(x)\cot(x)$.

Step4: Find the derivative of $f(x)$

$f'(x)$ is the sum of the derivatives of $-2\cot(x)$ and $6x^{4}\csc(x)$. So $f'(x)=2\csc^{2}(x)+24x^{3}\csc(x)-6x^{4}\csc(x)\cot(x)$.

Answer:

$2\csc^{2}(x)+24x^{3}\csc(x)-6x^{4}\csc(x)\cot(x)$