homework5: problem 22 (1 point) differentiate $y = csc x(x+cot x)$. $y=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u = \csc x$ and $v=x + \cot x$.

Step2: Find $u'$

The derivative of $\csc x$ is $-\csc x\cot x$, so $u'=-\csc x\cot x$.

Step3: Find $v'$

The derivative of $x$ is $1$ and the derivative of $\cot x$ is $-\csc^{2}x$. So $v'=1-\csc^{2}x$.

Step4: Calculate $y'$

$y'=u'v + uv'=-\csc x\cot x(x + \cot x)+\csc x(1-\csc^{2}x)$. Expand the expression: [ \begin{align*} y'&=-\csc x\cot x\cdot x-\csc x\cot^{2}x+\csc x-\csc^{3}x\ &=-\csc x(x\cot x+\cot^{2}x + \csc^{2}x - 1) \end{align*} ] Since $\csc^{2}x=1 + \cot^{2}x$, then $\csc^{2}x-1=\cot^{2}x$. [ \begin{align*} y'&=-\csc x(x\cot x+\cot^{2}x+\cot^{2}x)\ &=-\csc x(x\cot x + 2\cot^{2}x)\ &=-\csc x\cot x(x + 2\cot x) \end{align*} ]

Answer:

$-\csc x\cot x(x + 2\cot x)$