clara1223
  • clara1223
find f''(x) of f(x)=5cos(x^3)
Mathematics
katieb
  • katieb
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clara1223
  • clara1223
\[f''(x)\] of \[f(x)=5\cos(x ^{3})\]
zepdrix
  • zepdrix
Second derivative? Oh boy that'll be a bit of a doozy :o
zepdrix
  • zepdrix
Did you find \(\large\rm f'(x)\) yet?

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clara1223
  • clara1223
I'm not sure what to do with the x^3 inside
zepdrix
  • zepdrix
Chain... rule :o
zepdrix
  • zepdrix
So you expect that it should give you `something like this`\[\large\rm \frac{d}{dx}5\cos(x^3)=-5\sin(x^3)\]
zepdrix
  • zepdrix
But chain rule tells us to multiply by the derivative of the inner function.\[\large\rm \frac{d}{dx}5\cos(x^3)=-5\sin(x^3)\cdot\color{royalblue}{(x^3)'}\]
zepdrix
  • zepdrix
\[\large\rm \frac{d}{dx}5\cos(x^3)=-5\sin(x^3)\cdot\color{orangered}{(3x^2)}\]Ya?
clara1223
  • clara1223
ya so we have \[-15x ^{2}\sin(x ^{3})\] correct?
zepdrix
  • zepdrix
Looks good. Now things get a little more complicated. Looks like product rule from here, ya?
clara1223
  • clara1223
ya...
zepdrix
  • zepdrix
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.
clara1223
  • clara1223
i know that (-15x^2)' is -30x but what is (sin(x^3))'?
zepdrix
  • zepdrix
You should be able to do this one :) We just figured out what (cos(x^3))' is. This one is very very similar.
zepdrix
  • zepdrix
Claraaaaaaaaaaa! Gotta learn that chain rule!! MOAR PRACTICE! :)
clara1223
  • clara1223
OK so \[(\sin(x ^{3}))'=3x ^{2}\cos(x ^{3})\] yes?
zepdrix
  • zepdrix
Good :)
zepdrix
  • zepdrix
\[\large\rm \frac{d}{dx}-15x^2 \sin(x^3)=\color{royalblue}{(-15x^2)'}\sin(x^3)-15x^3\color{royalblue}{\left(\sin(x^3)\right)'}\] \[\large\rm \frac{d}{dx}-15x^2 \sin(x^3)=\color{orangered}{(-30x)}\sin(x^3)-15x^3\color{orangered}{\left(3x^2\cos(x^3)\right)}\]And then maybe combine things a lil bit.
clara1223
  • clara1223
\[-30x \sin (x ^{3})-45x ^{5}\cos(x ^{3})\] yes?
zepdrix
  • zepdrix
Woops! I made a boo boo somewhere!
clara1223
  • clara1223
where?
zepdrix
  • zepdrix
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?
anonymous
  • anonymous
zepdrix how do you get those colors for your latex equations?
anonymous
  • anonymous
dont mean to distract but I am just curious
ShadowLegendX
  • ShadowLegendX
@PlasmaFuzer Read a LaTeX tutorial?
zepdrix
  • zepdrix
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
zepdrix
  • zepdrix
example: \color{red}{x^2} you would put this manually into the equation tool \(\color{red}{x^2}\)
anonymous
  • anonymous
Ahh cool thank you... yea I use latex but never with colors but that is handy I will try and use it :D
clara1223
  • clara1223
@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?
zepdrix
  • zepdrix
@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.
zepdrix
  • zepdrix
It was a -15x^2, ya? Not -15x^3. I musta misclicked the power lol
anonymous
  • anonymous
Nice!!! thank you again :D :D anyways Ill stop distracting sorry clara cya both later
anonymous
  • anonymous
still relatively new to site... figuring things out
clara1223
  • clara1223
Oh! I got -15x^2, didn't even notice
zepdrix
  • zepdrix
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? :)
clara1223
  • clara1223
\[-30x \sin (x ^{3})-45x ^{4}\cos (x ^{3})\] right? because when you multiply powers you add them?
zepdrix
  • zepdrix
sec :) just double checking our work really quick
zepdrix
  • zepdrix
yayyy good job \c:/
clara1223
  • clara1223
I just multiplied the -15x^2 and the 3x^2 in front of the cos to get -45x^4
clara1223
  • clara1223
That's the most simplified it's gonna get right?
zepdrix
  • zepdrix
You could pull an x out of each term, but again, unnecessary :) Ya that's a good final answer.
clara1223
  • clara1223
Thank You!! you have been so helpful. I have 3 more questions tonight, hope you can help :)
zepdrix
  • zepdrix
Calc is so much fun isn't it claire bear? :O

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