anonymous
  • anonymous
If f(x)=x^3 cos3x then f '(x)=???
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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Psymon
  • Psymon
This would require the product rule, which would be: \[f'(x)g(x) + f(x)g'(x)\] Basically, the derivative of the first portion, the x^3, multiplied by simply cos3x. Then + x^3 times the derivative of cos3x. If I asked you to, separately, give me the derivative of x^3 and the derivative of cos3x could you tell me?
anonymous
  • anonymous
2x^2 and -sin3x?
Psymon
  • Psymon
Well, lets look at the first one, x^3. So the normal power rule for derivatives is: \[nx ^{n-1}\]Where n is the original power. SO with x^3, the original power is 3. So this formula tells us we want to bring the 3 down in front and then reduce the power by 1. Doing that gives us 3x^2. Do you see why?

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anonymous
  • anonymous
Yes, typed 2 by accident sorry. Is the derivative of cos3x right? Im not too sure if its correct.
Psymon
  • Psymon
No, its not actually. Almost, though. With trig function, they pretty much are always chain rules. The reason for this is all trig functions have 2 layers. They have the trigfunction itself, sin, cos, tan, etc, and they also have the inner angle. Because all trig functions or chain rules, we have to apply regular chain rule methods. So the chain rule is when we take all our layers, or inner functions if you prefer, and multiply all their derivatives. So what you did with cos3x becoming -sin3x is correct, but we have to then take into account the inner angle. So we would also have to multiply by the derivative of 3x, which is 3. So together the derivative of cos3x is -3sin3x. Make sense?
anonymous
  • anonymous
Yes, so then f'(x) would be the 3x^2 multiple -3sin3x
Psymon
  • Psymon
Well we still have to use theformula above. We multiply the derivative of the first, 3x^2 with simply cos3x. Then + x^3 with the derivative of cos3x, which is -3sin3x. So that would give us this: \[3x ^{2}\cos3x-3x ^{3}\sin3x \]Or if you had to simplify ever so slightly: \[3x ^{2}(\cos3x-xsin3x)\]
anonymous
  • anonymous
The formula i used was the one you had but divided by (g(x))^2. What is the difference between the two?
Psymon
  • Psymon
Thats for quotient rule. You would only use that if it was division, like if the problem was \[\frac{ x ^{3} }{ \cos3x } \] But the way you had it written it looked like a multiplication. Was it supposed to be division?
anonymous
  • anonymous
But the top is subtract, f′(x)g(x)-f(x)g′(x)
anonymous
  • anonymous
Ohh i got it now, thanks a lot :D
Psymon
  • Psymon
Lol, yeah, sure xD

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