find the derivative of the following function. y = x^2 csc x\n\frac{dy}{dx}=square

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Here, $u = x^{2}$ and $v=\csc x$.

Step2: Find $u^\prime$

Differentiate $u = x^{2}$ with respect to $x$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we get $u^\prime=\frac{d}{dx}(x^{2}) = 2x$.

Step3: Find $v^\prime$

Differentiate $v=\csc x$ with respect to $x$. The derivative of $\csc x$ is $-\csc x\cot x$, so $v^\prime=-\csc x\cot x$.

Step4: Substitute $u$, $u^\prime$, $v$, and $v^\prime$ into the product - rule

$y^\prime=u^\prime v+uv^\prime=(2x)\csc x+(x^{2})(-\csc x\cot x)$.

Step5: Simplify the expression

$y^\prime = 2x\csc x - x^{2}\csc x\cot x=\csc x(2x - x^{2}\cot x)$.

Answer:

$2x\csc x - x^{2}\csc x\cot x$