find y if y = 8csc x. y = □

Explanation:

Step1: Find the first - derivative

The derivative of $\csc x$ is $-\csc x\cot x$. Using the constant - multiple rule, if $y = 8\csc x$, then $y'=8\times(-\csc x\cot x)=- 8\csc x\cot x$.

Step2: Find the second - derivative

Use the product rule $(uv)' = u'v+uv'$, where $u=-8\csc x$ and $v = \cot x$. The derivative of $\csc x$ is $-\csc x\cot x$ and the derivative of $\cot x$ is $-\csc^{2}x$. $u'=-8\times(-\csc x\cot x)=8\csc x\cot x$, $v = \cot x$, $u=-8\csc x$, $v'=-\csc^{2}x$. $y''=u'v + uv'=(8\csc x\cot x)\cot x+(-8\csc x)(-\csc^{2}x)$. $y''=8\csc x\cot^{2}x + 8\csc^{3}x$. Since $\cot^{2}x=\csc^{2}x - 1$, we can also write $y''=8\csc x(\csc^{2}x - 1)+8\csc^{3}x$. Expand: $y''=8\csc^{3}x-8\csc x + 8\csc^{3}x=8(2\csc^{3}x-\csc x)$.

Answer:

$8(2\csc^{3}x-\csc x)$