homework4: problem 4\n(1 point)\ndifferentiate ( y=csc x(x+cot x) ).\n( y^{prime}= ) ( x csc (x) cot (x)-csc…
homework4: problem 4\n(1 point)\ndifferentiate ( y=csc x(x+cot x) ).\n( y^{prime}= ) ( x csc (x) cot (x)-csc (x) cot ^{2}(x)+csc (x)-csc ^{3}(x) )\npreview my answers submit answers\nyou have attempted this problem 1 time.\nyour overall recorded score is ( 0 % ).\nyou have unlimited attempts remaining.
Answer
Explanation:
Step1: Apply the 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). First, find (u') and (v'). The derivative of (\csc x) is (-\csc x\cot x) (i.e., (u'=-\csc x\cot x)), and the derivative of (v=x+\cot x) is (v' = 1-\csc^{2}x) (since the derivative of (x) is (1) and the derivative of (\cot x=-\csc^{2}x)).
Step2: Substitute into the product rule formula
[ \begin{align*} y'&=u'v+uv'\ &=(-\csc x\cot x)(x + \cot x)+\csc x(1-\csc^{2}x)\ &=-x\csc x\cot x-\csc x\cot^{2}x+\csc x-\csc^{3}x \end{align*} ]
Answer:
(y'=-x\csc x\cot x-\csc x\cot^{2}x+\csc x-\csc^{3}x)