clara1223
  • clara1223
find f'(x) of f(x)=sin(2x)(sqrt(1+cos(5x)))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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misty1212
  • misty1212
HI!!
misty1212
  • misty1212
product plus chain rule \[(fg)'=f'g+g'f\] with \[f(x)=\sin(x), f'(x)=\cos(x), g(x)=\sqrt{1+\cos(5x)}\] \[g'(x)=\frac{-5\sin(5x)}{2\sqrt{1+\cos(5x)}}\]
clara1223
  • clara1223
what happened to the 2x inside sin(x)?

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misty1212
  • misty1212
oops \[f(x)=\sin(2x), f'(x)=2\cos(2x)\] my bad
clara1223
  • clara1223
How would I apply the chain rule to that? The sqrt makes it pretty messy.
clara1223
  • clara1223
@misty1212 ?
misty1212
  • misty1212
\[\left(\sqrt{f}\right)'=\frac{f'}{2\sqrt{f}}\]
misty1212
  • misty1212
which should explain \[g'(x)=\frac{-5\sin(5x)}{2\sqrt{1+\cos(5x)}}\]
clara1223
  • clara1223
That makes sense. I'm just not sure what to do from there. From the product rule I got 2cos(2x)(sqrt(1+cos(5x)))+sin(2x)((-5sin(5x))/2(sqrt(1+cos(5x))))
misty1212
  • misty1212
leave it
clara1223
  • clara1223
that's my final answer?
misty1212
  • misty1212
it is a silly made up question anyway what else can you do?
misty1212
  • misty1212
you sure as hell don't want to actually add them although you could if you had like an extra half hour to waste
clara1223
  • clara1223
Haha definitely not. Still got 6 more questions on this homework sheet due tomorrow.
misty1212
  • misty1212
best get busy
misty1212
  • misty1212
lol not that "get busy"

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