Question

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

Answer 1

@zepdrix

Answer 2

@dan815 @e.mccormick @Emily778 @Euler271 @Hero @jim_thompson5910 @Luigi0210 @modphysnoob @primeralph

Answer 3

Chain rule:

Answer 4

F(g(x))=F'(g(x))*G'(x)

Answer 5

So the question is: is my statement in the first post correct , if no why not?

Answer 6

Now, you have three functions going on there. Do you see that part?

Answer 7

Honestly I have no idea what's going on , the most difficult thing I ever had to differentiate

Answer 8

@zepdrix

Answer 9

Lets rewrite it.\[\tan^3(\sqrt{x})=[tan(x^{\frac{1}{2}})]^3\]Does that help you see three functions?

Answer 10

hold on

Answer 11

that last part is just product rule, so not that hard to do once you know the derivative of this part.

Answer 12

uhm I dunno how you define a function, I cant count it

Answer 13

tan is a function, x with this power is a function, the third?

Answer 14

Any time x has something done to it, it can be seen as a new function.

Answer 15

The whole thing to the 3rd power is the 3rd.

Answer 16

alright and then?

Answer 17

So, you ever seen the chain rule applied multiple times?

Answer 18

just twice

Answer 19

now it needs three times?

Answer 20

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.

Answer 21

that gets me (x^(-1/2)tan^3(sqrt(x)))/2 + sqrt(x)3(tan^2(sqrt(x))2tan(sqrt(x))

?

Answer 22

\(d/dx~ h(f(n))=h'(f(n))f'(n)\)
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:
\(d/dx~ h(f(g(x)))=h'(f(g(x)))f'(g(x))g'(x)\)

Answer 23

\[\large y'=\color{orangered}{(\frac{1}{2}x^{-1/2})}\tan^3\sqrt{x}+\sqrt{x}\color{royalblue}{(\tan^3\sqrt{x})'}\]Hmm, your first derivative looks correct, let's see if we can figure out what's going on with the other one.

Answer 24

ok

Answer 25

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.

Answer 26

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

Answer 27

oh my bad hehe c:

Answer 28

np. It all works. Not like we are contradicting each other. LOL. I have had that happen.

Answer 29

I see how it goes now, lets see what I can do about this

Answer 30

go team go! (Well, a team of 1....)

Answer 31

(x(-1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x(-1/2))/2

oh boy

Answer 32

the second "(" count it as a "^"

Answer 33

(x^(-1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x(-1/2))/2

Answer 34

(x^(-1/2)/2)*tan^3(sqrt(x))+(3sqrt(x)tan^2(sqrt(x))sec^2(sqrt(x))x^(-1/2))/2

Answer 35

\[\large y'=\color{orangered}{(\frac{1}{2}x^{-1/2})}\tan^3\sqrt{x}+\sqrt{x}\color{orangered}{(\large 3\tan^2\sqrt x)(\sec^2\sqrt x)(\frac{1}{2}x^{-1/2})}\]
yah it looks like you did it correctly. :) should probably be simplified down a smidge though.

Answer 36

ty bro