Question

question
find the derivative of $f(x)=7csc(x)+\frac{1}{x}$.
provide your answer below:
$f(x)=square$

Explanation:

Step1: Recall derivative rules

The derivative of a sum of functions is the sum of their derivatives. So $f'(x)=(7\csc(x))'+(\frac{1}{x})'$.

Step2: Differentiate $7\csc(x)$

The derivative of $\csc(x)$ is $-\csc(x)\cot(x)$. Using the constant - multiple rule $(cf(x))' = cf'(x)$, the derivative of $7\csc(x)$ is $7(-\csc(x)\cot(x))=-7\csc(x)\cot(x)$.

Step3: Differentiate $\frac{1}{x}$

Rewrite $\frac{1}{x}$ as $x^{-1}$. Using the power - rule $(x^n)'=nx^{n - 1}$, the derivative of $x^{-1}$ is $-1\times x^{-1 - 1}=-x^{-2}=-\frac{1}{x^{2}}$.

Step4: Combine the derivatives

$f'(x)=-7\csc(x)\cot(x)-\frac{1}{x^{2}}$.

Answer:

$-7\csc(x)\cot(x)-\frac{1}{x^{2}}$