Question

Need help! Can anyone please help me? It would be much appreciated. Please view the attached photo. Will mark as brainliest!

Answer 1

Answer:

  $\text{D)} \quad y'=-10\csc{x}\cot{x}+\sec^2{x}$

Step-by-step explanation:

  $y=\dfrac{10}{\sin{x}}+\dfrac{1}{\cot{x}}=10\sin{(x)}^{-1}+\sin{(x)}\cos{(x)}^{-1}\\\\y'=-10\sin{(x)}^{-2}\cos{(x)}+\cos{(x)}\cos{(x)}^{-1}+\sin{(x)}^2\cos{{x}}^{-2}\\\\=-10\dfrac{1}{\sin{x}}\cdot\dfrac{\cos{x}}{\sin{x}}+1+\left(\dfrac{\sin{x}}{\cos{x}}\right)^2\\\\=-10\csc{x}\cot{x}+\sec^2{x}$

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The product rule for derivatives applies.

  y = fg   ⇒   y' = f'g +fg'