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anonymous

  • one year ago

How do I find the derivative of...

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  1. anonymous
    • one year ago
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    \[f(x)=\frac{ 5 }{ (2x)^3 }+2cosx\]

  2. anonymous
    • one year ago
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    @zepdrix ?

  3. zepdrix
    • one year ago
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    A rule of exponents let's us write it like this: \(\large\rm f(x)=5(2x)^{-3}+2\cos x\)

  4. zepdrix
    • one year ago
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    Differentiating the first term shouldn't be too bad, power rule into chain rule, ya?

  5. anonymous
    • one year ago
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    \[-15(2x)^{-4}\]

  6. anonymous
    • one year ago
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    I am horrible with trigonometric identities though, so I'm not sure what to do with the cos part.

  7. zepdrix
    • one year ago
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    Woops:\[\large\rm \color{royalblue}{\left[5(2x)^{-3}\right]'}=-3\cdot5(2x)^{-4}\cdot\color{royalblue}{(2x)'}\]Forgot your chain rule I think.

  8. zepdrix
    • one year ago
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    \[\large\rm =-3\cdot5(2x)^{-4}\cdot(2)\]

  9. anonymous
    • one year ago
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    I'm a bit unfamiliar with the rule... So you would take the 2 from (2x) and multiply it by the rest?

  10. zepdrix
    • one year ago
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    You learn these tricks early on: Power Rule Product Rule Quotient Rule Chain Rule Chain Rule is the most difficult of these to master, by far. You'll need to do a lot lot lot of practice problems to feel comfortable with it.

  11. zepdrix
    • one year ago
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    You're always multiplying by the derivative of the `inner function`. So yes, for this problem, the inner function is 2x. We have to multiply by the derivative of 2x on outside like that.

  12. zepdrix
    • one year ago
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    Another example: \(\large\rm 5(x^2+3)^5\) I apply the power rule to outermost function which is \(\large\rm 5(\qquad\quad)^5\) giving me \(\large\rm 5\cdot5(\qquad\quad)^4\) Chain rule tells me that I have to multiply by the derivative of the inner function.\[\large\rm 5\cdot5(\qquad\quad)^4\cdot\left(\qquad\quad\right)'\]Where ( ) is whatever was inside of the outerfunction that we had. The prime says to take a derivative. \[\large\rm 5\cdot5(\qquad\quad)^4\cdot (2x+0)\]So I multiplied by the derivative of x^2+3, the inner function. So the final result is:\(\large\rm 5\cdot5(x^2+3)^4(2x)\) Just a silly example :O I hope the empty brackets didn't make it MORE confusing lol

  13. zepdrix
    • one year ago
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    For the trig stuff.. honestly, just memorize them for now. Sine and cosine are very cyclical. When you differentiate sine a number of times, it follows this pattern: \(\large\rm \sin x\quad\to\quad \cos x\quad\to\quad -\sin x\quad\to\quad-\cos x\quad\to\quad\sin x\) So if you differentiate sin x, you get cos x. If you differentiate cos x, you get -sin x. If you differentiate sin x `four times`, you get right back to sin x! :)

  14. anonymous
    • one year ago
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    Hey I'm back. Sorry, my internet gave out for a bit. It tends to do that every now and then.

  15. zepdrix
    • one year ago
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    And keep in mind that constant coefficients have effect on the differentiation process. Just carry the 2 along for the ride.\[\large\rm 2\sin x\quad\to\quad 2\cos x\quad\to\quad -2\sin x\quad\to\quad-2\cos x\quad\to\quad2\sin x\]

  16. zepdrix
    • one year ago
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    have no effect*

  17. anonymous
    • one year ago
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    I see how in your "Another example" you got the 2x from x^2, but why did the 3 turn into a zero?

  18. zepdrix
    • one year ago
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    Differentiation is all about measuring "change". Something that is `constant` by definition, does not change. So when I'm taking a derivative, this "dx" represents an immeasurably small change in x. So when d/dx looks at the 3, he's asking the question "when x changes a very small amount, how much does 3 change?" And the answer is 0 amount. It's rate of change is 0. If you prefer, you can think of it in terms of power rule:\[\large\rm \frac{d}{dx}3\cdot\color{orangered}{1}=\frac{d}{dx}3\cdot\color{orangered}{x^0}=3\cdot0x^{-1}=0\]

  19. zepdrix
    • one year ago
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    Recall that anything to the zero power is just 1. \(\large\rm a^0=1,\qquad \forall a\ne0\). So I used this clever idea to rewrite 1 as x to the 0.

  20. anonymous
    • one year ago
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    So any number besides 3 would also turn into zero as well?

  21. zepdrix
    • one year ago
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    Yes. And later on, they might try to trick you and give you "fancy numbers" as I like to call them.

  22. zepdrix
    • one year ago
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    \[\large\rm \left(e^{4\pi}\right)'=0\]There is no variable! The whole thing is just a constant!

  23. anonymous
    • one year ago
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    Ah, okay. That seems simple enough.

  24. zepdrix
    • one year ago
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    Polynomials will follow this progression as you differentiate them:\[\large\rm cubic\quad\to\quad square\quad\to\quad linear \quad\to\quad constant\quad\to\quad 0\]

  25. zepdrix
    • one year ago
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    That's just an illustration of the power rule ^ Obviously higher power follow it as well :)

  26. zepdrix
    • one year ago
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    Bahh I gotta run to the grocery store before it closes :O You got this figured out? XD

  27. anonymous
    • one year ago
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    I believe so. Oh quick question. I just convert cosine to -sine right?

  28. anonymous
    • one year ago
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    If that will take too much time to explain, then don't worry about it. I don't want to be the cause of you not having food. :)

  29. zepdrix
    • one year ago
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    Good! :D And the 2 comes along for the ride of course.\[\large\rm f(x)=5(2x)^{-3}+2\cos x\]\[\large\rm f'(x)=-3\cdot5(2x)^{-4}(2)-2\sin x\]

  30. anonymous
    • one year ago
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    Thank you so much!

  31. zepdrix
    • one year ago
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    yay team \c:/

  32. anonymous
    • one year ago
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    You're better at explaining this than my textbook or teacher.

  33. zepdrix
    • one year ago
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    Ehh it's easy to say that when you're getting 1 on 1 attention XD lol Gotta give teacher just a little bit of slack haha

  34. anonymous
    • one year ago
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    Well, I have in the past but eh. I guess I just have a different learning style. Good night! :)

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