Problem of the Week

Updated at May 5, 2014 5:51 PM

For this week we've brought you this calculus problem.

How can we find the derivative of 1sin⁡2x\frac{1}{\sin^{2}x}?

Here are the steps:



ddx1sin⁡2x\frac{d}{dx} \frac{1}{\sin^{2}x}

1
Use Chain Rule on ddx1sin⁡2x\frac{d}{dx} \frac{1}{\sin^{2}x}. Let u=sin⁡xu=\sin{x}. Use Power Rule: dduun=nun−1\frac{d}{du} {u}^{n}=n{u}^{n-1}.
−2sin⁡3x(ddxsin⁡x)-\frac{2}{\sin^{3}x}(\frac{d}{dx} \sin{x})

2
Use Trigonometric Differentiation: the derivative of sin⁡x\sin{x} is cos⁡x\cos{x}.
−2cos⁡xsin⁡3x-\frac{2\cos{x}}{\sin^{3}x}

Done
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