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

- clara1223

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

- katieb

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- clara1223

\[f''(x)\] of \[f(x)=5\cos(x ^{3})\]

- zepdrix

Second derivative?
Oh boy that'll be a bit of a doozy :o

- zepdrix

Did you find \(\large\rm f'(x)\) yet?

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## More answers

- clara1223

I'm not sure what to do with the x^3 inside

- zepdrix

Chain... rule :o

- 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

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

\[\large\rm \frac{d}{dx}5\cos(x^3)=-5\sin(x^3)\cdot\color{orangered}{(3x^2)}\]Ya?

- clara1223

ya so we have \[-15x ^{2}\sin(x ^{3})\] correct?

- zepdrix

Looks good.
Now things get a little more complicated.
Looks like product rule from here, ya?

- clara1223

ya...

- 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

i know that (-15x^2)' is -30x but what is (sin(x^3))'?

- 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

Claraaaaaaaaaaa!
Gotta learn that chain rule!! MOAR PRACTICE! :)

- clara1223

OK so \[(\sin(x ^{3}))'=3x ^{2}\cos(x ^{3})\] yes?

- zepdrix

Good :)

- 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

\[-30x \sin (x ^{3})-45x ^{5}\cos(x ^{3})\] yes?

- zepdrix

Woops! I made a boo boo somewhere!

- clara1223

where?

- 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

zepdrix how do you get those colors for your latex equations?

- anonymous

dont mean to distract but I am just curious

- ShadowLegendX

@PlasmaFuzer Read a LaTeX tutorial?

- 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

example:
\color{red}{x^2}
you would put this manually into the equation tool
\(\color{red}{x^2}\)

- 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

@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

@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

It was a -15x^2, ya?
Not -15x^3.
I musta misclicked the power lol

- anonymous

Nice!!! thank you again :D :D anyways Ill stop distracting sorry clara cya both later

- anonymous

still relatively new to site... figuring things out

- clara1223

Oh! I got -15x^2, didn't even notice

- 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

\[-30x \sin (x ^{3})-45x ^{4}\cos (x ^{3})\] right? because when you multiply powers you add them?

- zepdrix

sec :) just double checking our work really quick

- zepdrix

yayyy good job \c:/

- clara1223

I just multiplied the -15x^2 and the 3x^2 in front of the cos to get -45x^4

- clara1223

That's the most simplified it's gonna get right?

- zepdrix

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

- clara1223

Thank You!! you have been so helpful. I have 3 more questions tonight, hope you can help :)

- zepdrix

Calc is so much fun isn't it claire bear? :O

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