anonymous
  • anonymous
Differential calculus help whats the diff when differentiating sin x sin^2 x sin 2x
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
btw and 2sin x
Isaiah.Feynman
  • Isaiah.Feynman
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Psymon
  • Psymon
Oh, derivatives O.o

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Isaiah.Feynman
  • Isaiah.Feynman
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anonymous
  • anonymous
srry wrong type lol edited now
Psymon
  • Psymon
Oh, so it was derivatives.
Psymon
  • Psymon
Yeah, okay, not integration, differentiation, haha.
Psymon
  • Psymon
Well, the derviative of sin x is simply cosx. That one is just straightforward. The other two are chain rule. We have to think of each function as broken up into layers. For example, sin^2(x). As was mentioned, sin^2(x) is the same as (sinx)^2. Writing it in this manner makes the chain rule a little more obvious. So this is what we do: (sinx)^2 can be thought of as having two layers: \[(--)^{2}\] \[sinx \] The outer layer is simply the exponential portion and the inner layer is sinx. When we do the chain rule, we take the derivative of each layer and multiply the results. We DO NOT alter the inside of each layer. So starting with the exponent layer. Regardless of what's inside, the power rule says: \[\frac{ d }{ dx }(--)^{2}=2(--)^{1} \] THis is the derivative of the first layer. Now we take the derivative of the second layer. \[\frac{ d }{ dx }sinx = cosx \]So now we found our two layers and took the derivative of each layer. FInally, we multiply the results, which gives us: \[2(--)^{1}(cosx)= 2(sinx)(cosx)\]For that last part, I just plugged back in what was originally in the layer because, as I said, the inner part of the layer does not get altered. Finally we have sin(2x). this is also a chain rule of 2 layers. \[\sin(--)\]and \[2x \] As before, we take the derivative of each layer and multiply the results. So: \[\frac{ d }{ dx }\sin(--)=\cos(--)\] \[\frac{ d }{ dx }2x=2 \] Now that I have my two derivatives, I multiply the results to give us: \[2\cos(--) = 2\cos(2x)\]Once again, making sure to plug back in what was originally inside of the layer.
anonymous
  • anonymous
cool i realize i was confused about the structure of the function lol.
Psymon
  • Psymon
Well hopefully it helps to see it in that way :3 And if you got any other examples Im glad to show ya :3

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