Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (clara1223):

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

OpenStudy (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?

OpenStudy (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?

OpenStudy (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?

OpenStudy (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.

OpenStudy (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! :)

OpenStudy (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.

OpenStudy (clara1223):

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

zepdrix (zepdrix):

Woops! I made a boo boo somewhere!

OpenStudy (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?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

dont mean to distract but I am just curious

OpenStudy (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 \(\color{red}{x^2}\)

OpenStudy (anonymous):

Ahh cool thank you... yea I use latex but never with colors but that is handy I will try and use it :D

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

still relatively new to site... figuring things out

OpenStudy (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? :)

OpenStudy (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:/

OpenStudy (clara1223):

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

OpenStudy (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.

OpenStudy (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

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!