I’m in AP Calculus AB. Can you help me solve this showing steps?I’m not exactly sure how to do this.

We need to find the derivative of a function, but if we look at the function is a quotient. Then we can use the rule for the derivative of a quotient:
[tex]\begin{gathered} f(x)=\frac{g(x)}{h(x)} \\ \text{Then,} \\ \frac{df}{dx}=\frac{g^{\prime}h-h^{\prime}g}{h^2} \end{gathered}[/tex]In this case, we have:
[tex]f(x)=\frac{\csc (x)}{x^2}[/tex]We can call:
[tex]\begin{gathered} g(x)=\csc (x) \\ h(x)=x^2 \end{gathered}[/tex]Now we apply the rule:
[tex]\frac{df}{dx}=\frac{\frac{dg}{dx}h-\frac{dh}{dx}g}{h^2}[/tex]We know that the derivative of h(x) = x^2 is 2x, and the derivative of g(x) = csc(x) is [-cot(x)csc(x)]
Replacing the values:
[tex]\frac{df}{dx}=\frac{-\cot (x)\csc (x)\cdot x^2-2x\cdot\csc (x)}{(x^2)^2}[/tex]Simplifying:
We can factor out csc(x)*x in the numerator and solve the square in the denominator
[tex]\frac{df}{dx}=\frac{x\csc (x)(-x\cot (x)-2)}{x^4}[/tex]Finally, we can factor the (-1) inside the parentheses and cancel out the x in the numerator:
[tex]\frac{df}{dx}=-\frac{\csc (x)(x\cot (x)+2)}{x^3}[/tex]And this is the final answer.