Problem of the Week

Updated at Oct 28, 2013 9:02 AM

How can we solve for the derivative of sin⁡xcos⁡2x\frac{\sin{x}}{\cos^{2}x}?

Below is the solution.



ddxsin⁡xcos⁡2x\frac{d}{dx} \frac{\sin{x}}{\cos^{2}x}

1
Use Quotient Rule to find the derivative of sin⁡xcos⁡2x\frac{\sin{x}}{\cos^{2}x}. The quotient rule states that (fg)′=f′g−fg′g2(\frac{f}{g})'=\frac{f'g-fg'}{{g}^{2}}.
cos⁡2x(ddxsin⁡x)−sin⁡x(ddxcos⁡2x)cos⁡4x\frac{\cos^{2}x(\frac{d}{dx} \sin{x})-\sin{x}(\frac{d}{dx} \cos^{2}x)}{\cos^{4}x}

2
Use Trigonometric Differentiation: the derivative of sin⁡x\sin{x} is cos⁡x\cos{x}.
cos⁡3x−sin⁡x(ddxcos⁡2x)cos⁡4x\frac{\cos^{3}x-\sin{x}(\frac{d}{dx} \cos^{2}x)}{\cos^{4}x}

3
Use Chain Rule on ddxcos⁡2x\frac{d}{dx} \cos^{2}x. Let u=cos⁡xu=\cos{x}. Use Power Rule: dduun=nun−1\frac{d}{du} {u}^{n}=n{u}^{n-1}.
cos⁡3x−sin⁡x×2cos⁡x(ddxcos⁡x)cos⁡4x\frac{\cos^{3}x-\sin{x}\times 2\cos{x}(\frac{d}{dx} \cos{x})}{\cos^{4}x}

4
Use Trigonometric Differentiation: the derivative of cos⁡x\cos{x} is −sin⁡x-\sin{x}.
1cos⁡x+2sin⁡2xcos⁡3x\frac{1}{\cos{x}}+\frac{2\sin^{2}x}{\cos^{3}x}

Done
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