Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Christos
Group Title
the derivative of tan^3(sqrt(x))
is 3tan^2(sqrt(x))2tan(sqrt(x))
?
I am asking in an attempt to actually find this derivative http://screencast.com/t/1uOEBG1WH
 one year ago
 one year ago
Christos Group Title
the derivative of tan^3(sqrt(x)) is 3tan^2(sqrt(x))2tan(sqrt(x)) ? I am asking in an attempt to actually find this derivative http://screencast.com/t/1uOEBG1WH
 one year ago
 one year ago

This Question is Closed

Christos Group TitleBest ResponseYou've already chosen the best response.0
@dan815 @e.mccormick @Emily778 @Euler271 @Hero @jim_thompson5910 @Luigi0210 @modphysnoob @primeralph
 one year ago

Luigi0210 Group TitleBest ResponseYou've already chosen the best response.0
Chain rule:
 one year ago

Luigi0210 Group TitleBest ResponseYou've already chosen the best response.0
F(g(x))=F'(g(x))*G'(x)
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
So the question is: is my statement in the first post correct , if no why not?
 one year ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
Now, you have three functions going on there. Do you see that part?
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
Honestly I have no idea what's going on , the most difficult thing I ever had to differentiate
 one year ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
Lets rewrite it.\[\tan^3(\sqrt{x})=[tan(x^{\frac{1}{2}})]^3\]Does that help you see three functions?
 one year ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
that last part is just product rule, so not that hard to do once you know the derivative of this part.
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
uhm I dunno how you define a function, I cant count it
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
tan is a function, x with this power is a function, the third?
 one year ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
Any time x has something done to it, it can be seen as a new function.
 one year ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
The whole thing to the 3rd power is the 3rd.
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
alright and then?
 one year ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
So, you ever seen the chain rule applied multiple times?
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
just twice
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
now it needs three times?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
The setup will be the same as the last problem ms christos :o \[\large y=\sqrt{x}\tan^3\sqrt{x}\] \[\large y'=\color{royalblue}{(\sqrt{x})'}\tan^3\sqrt{x}+\sqrt{x}\color{royalblue}{(\tan^3\sqrt{x})'}\] Remember how we did that? Product rule setup.
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
that gets me (x^(1/2)tan^3(sqrt(x)))/2 + sqrt(x)3(tan^2(sqrt(x))2tan(sqrt(x)) ?
 one year ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
\(d/dx~ h(f(n))=h'(f(n))f'(n)\) But what if n=g(x)? Then f'(n) is another chain rule! \(d/dx~ f(g(x))=f'(g(x))g'(x)\) When I put the two together: \(d/dx~ h(f(g(x)))=h'(f(g(x)))f'(g(x))g'(x)\)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\large y'=\color{orangered}{(\frac{1}{2}x^{1/2})}\tan^3\sqrt{x}+\sqrt{x}\color{royalblue}{(\tan^3\sqrt{x})'}\]Hmm, your first derivative looks correct, let's see if we can figure out what's going on with the other one.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Let's just look at the blue term for now,\[\large \color{royalblue}{(\tan^3\sqrt x)'}\] So the outermost function is (something) cubed. Applying the power rule and chain rule gives us,\[\large 3(\tan^2\sqrt x)\color{royalblue}{(\tan\sqrt x)'}\]Three came down, giving us a 2 exponent, then we apply the chain rule. Remember the derivative of tangent?\[\large 3(\tan^2\sqrt x)(\sec^2\sqrt x)\color{royalblue}{(\sqrt x)'}\] Are you following my notation ok? The blue terms are ones that we need to take a derivative of. They keep showing up due to the chain rule.
 one year ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
A lot of people like to start at the inner most function, and work their way out. It is not a bad choice because it works well for multiple chain rules. Or work from outer in. However, as zepdrix points out, there is a problem with part of what you did for a derivative. What is the derivative of tangent? hmm.. he got it as I was typing in my word processor. LOL
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
oh my bad hehe c:
 one year ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
np. It all works. Not like we are contradicting each other. LOL. I have had that happen.
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
I see how it goes now, lets see what I can do about this
 one year ago

e.mccormick Group TitleBest ResponseYou've already chosen the best response.0
go team go! (Well, a team of 1....)
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
(x(1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x(1/2))/2 oh boy
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
the second "(" count it as a "^"
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
(x^(1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x(1/2))/2
 one year ago

Christos Group TitleBest ResponseYou've already chosen the best response.0
(x^(1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x^(1/2))/2
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
\[\large y'=\color{orangered}{(\frac{1}{2}x^{1/2})}\tan^3\sqrt{x}+\sqrt{x}\color{orangered}{(\large 3\tan^2\sqrt x)(\sec^2\sqrt x)(\frac{1}{2}x^{1/2})}\] yah it looks like you did it correctly. :) should probably be simplified down a smidge though.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.