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 \(\frac{1}{\sin^{2}x}\)?

Here are the steps:



\[\frac{d}{dx} \frac{1}{\sin^{2}x}\]

1
Use Chain Rule on \(\frac{d}{dx} \frac{1}{\sin^{2}x}\). Let \(u=\sin{x}\). Use Power Rule: \(\frac{d}{du} {u}^{n}=n{u}^{n-1}\).
\[-\frac{2}{\sin^{3}x}(\frac{d}{dx} \sin{x})\]

2
Use Trigonometric Differentiation: the derivative of \(\sin{x}\) is \(\cos{x}\).
\[-\frac{2\cos{x}}{\sin^{3}x}\]

Done
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