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Derivative of f(x) = sinxcosx

  • Thread starter Ryuk1990
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  • #1
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Homework Statement



f(x) = sinxcosx

Homework Equations



Product Rule: f(x)g(x) = f(x)Dg(x) + g(x)Df(x)

The Attempt at a Solution



I got to sinx(-sinx) + cosxcosx

The answer is supposed to be 0

Would the next step be turning it into -(sinx)2 + (cosx)2?
 

Answers and Replies

  • #2
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Your work is right.

It can't be zero, because otherwise, you could integrate it and you'd obtain a constant, and clearly sinxcosx is not a constant. It's actually (sin 2x)/2
 
  • #3
jbunniii
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Your answer is right (both forms).

Another way to solve this is to recognize that (as l'Hôpital said):

[tex]\sin x \cos x = \frac{1}{2} \sin 2x[/tex]

which has derivative

[tex]\cos 2x[/tex]

The answer looks different, but remembering one's trig identities pays dividends:

[tex]\cos 2x = \cos^2 x - \sin^2 x[/tex]
 

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