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clara1223

  • one year ago

find f''(x) of f(x)=5cos(x^3)

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  1. clara1223
    • one year ago
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    \[f''(x)\] of \[f(x)=5\cos(x ^{3})\]

  2. zepdrix
    • one year ago
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    Second derivative? Oh boy that'll be a bit of a doozy :o

  3. zepdrix
    • one year ago
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    Did you find \(\large\rm f'(x)\) yet?

  4. clara1223
    • one year ago
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    I'm not sure what to do with the x^3 inside

  5. zepdrix
    • one year ago
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    Chain... rule :o

  6. zepdrix
    • one year ago
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    So you expect that it should give you `something like this`\[\large\rm \frac{d}{dx}5\cos(x^3)=-5\sin(x^3)\]

  7. zepdrix
    • one year ago
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    But chain rule tells us to multiply by the derivative of the inner function.\[\large\rm \frac{d}{dx}5\cos(x^3)=-5\sin(x^3)\cdot\color{royalblue}{(x^3)'}\]

  8. zepdrix
    • one year ago
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    \[\large\rm \frac{d}{dx}5\cos(x^3)=-5\sin(x^3)\cdot\color{orangered}{(3x^2)}\]Ya?

  9. clara1223
    • one year ago
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    ya so we have \[-15x ^{2}\sin(x ^{3})\] correct?

  10. zepdrix
    • one year ago
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    Looks good. Now things get a little more complicated. Looks like product rule from here, ya?

  11. clara1223
    • one year ago
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    ya...

  12. zepdrix
    • one year ago
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    Here is how you would "set it up"\[\large\rm \frac{d}{dx}-15x^2 \sin(x^3)=\color{royalblue}{(-15x^2)'}\sin(x^3)-15x^3\color{royalblue}{\left(\sin(x^3)\right)'}\]And you need to differentiate the blue parts.

  13. clara1223
    • one year ago
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    i know that (-15x^2)' is -30x but what is (sin(x^3))'?

  14. zepdrix
    • one year ago
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    You should be able to do this one :) We just figured out what (cos(x^3))' is. This one is very very similar.

  15. zepdrix
    • one year ago
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    Claraaaaaaaaaaa! Gotta learn that chain rule!! MOAR PRACTICE! :)

  16. clara1223
    • one year ago
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    OK so \[(\sin(x ^{3}))'=3x ^{2}\cos(x ^{3})\] yes?

  17. zepdrix
    • one year ago
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    Good :)

  18. zepdrix
    • one year ago
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    \[\large\rm \frac{d}{dx}-15x^2 \sin(x^3)=\color{royalblue}{(-15x^2)'}\sin(x^3)-15x^3\color{royalblue}{\left(\sin(x^3)\right)'}\] \[\large\rm \frac{d}{dx}-15x^2 \sin(x^3)=\color{orangered}{(-30x)}\sin(x^3)-15x^3\color{orangered}{\left(3x^2\cos(x^3)\right)}\]And then maybe combine things a lil bit.

  19. clara1223
    • one year ago
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    \[-30x \sin (x ^{3})-45x ^{5}\cos(x ^{3})\] yes?

  20. zepdrix
    • one year ago
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    Woops! I made a boo boo somewhere!

  21. clara1223
    • one year ago
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    where?

  22. zepdrix
    • one year ago
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    When I did the setup:\[\large\rm \frac{d}{dx}\left[-15x^2 \sin(x^3)\right]=\color{royalblue}{(-15x^2)'}\sin(x^3)\color{red}{-15x^3}\color{royalblue}{\left(\sin(x^3)\right)'}\]Do you see anything wrong with this red part?

  23. anonymous
    • one year ago
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    zepdrix how do you get those colors for your latex equations?

  24. anonymous
    • one year ago
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    dont mean to distract but I am just curious

  25. ShadowLegendX
    • one year ago
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    @PlasmaFuzer Read a LaTeX tutorial?

  26. zepdrix
    • one year ago
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    The formatting language is LaTex You can google it and find out a lot of cool little features :) \color{specific color goes here}{your math goes here} for color

  27. zepdrix
    • one year ago
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    example: \color{red}{x^2} you would put this manually into the equation tool \(\color{red}{x^2}\)

  28. anonymous
    • one year ago
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    Ahh cool thank you... yea I use latex but never with colors but that is handy I will try and use it :D

  29. clara1223
    • one year ago
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    @zepdrix I don't see anything wrong with the red part. When I do the product rule I get the same thing... What am I missing?

  30. zepdrix
    • one year ago
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    @PlasmaFuzer Another cool thing you can do is: Right click someone's math code > Show Math As > Tex Commands and that will give you a pop up box of how I/someone entered it into the equation tool.

  31. zepdrix
    • one year ago
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    It was a -15x^2, ya? Not -15x^3. I musta misclicked the power lol

  32. anonymous
    • one year ago
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    Nice!!! thank you again :D :D anyways Ill stop distracting sorry clara cya both later

  33. anonymous
    • one year ago
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    still relatively new to site... figuring things out

  34. clara1223
    • one year ago
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    Oh! I got -15x^2, didn't even notice

  35. zepdrix
    • one year ago
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    So after differentiating, I should've had,\[\large\rm \frac{d}{dx}-15x^2 \sin(x^3)=\color{orangered}{(-30x)}\sin(x^3)-15x^2\color{orangered}{\left(3x^2\cos(x^3)\right)}\]So we don't end up with a 5th power over there, ya? :)

  36. clara1223
    • one year ago
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    \[-30x \sin (x ^{3})-45x ^{4}\cos (x ^{3})\] right? because when you multiply powers you add them?

  37. zepdrix
    • one year ago
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    sec :) just double checking our work really quick

  38. zepdrix
    • one year ago
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    yayyy good job \c:/

  39. clara1223
    • one year ago
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    I just multiplied the -15x^2 and the 3x^2 in front of the cos to get -45x^4

  40. clara1223
    • one year ago
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    That's the most simplified it's gonna get right?

  41. zepdrix
    • one year ago
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    You could pull an x out of each term, but again, unnecessary :) Ya that's a good final answer.

  42. clara1223
    • one year ago
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    Thank You!! you have been so helpful. I have 3 more questions tonight, hope you can help :)

  43. zepdrix
    • one year ago
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    Calc is so much fun isn't it claire bear? :O

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