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Christos
 3 years ago
Find derivative of
http://screencast.com/t/sXZqTgJKM
Just tell me 12 steps I will clip to them afterwards
Christos
 3 years ago
Find derivative of http://screencast.com/t/sXZqTgJKM Just tell me 12 steps I will clip to them afterwards

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abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0product rule within chain rule.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2\[\large y=x^3 \sin^2(5x)\]We start by setting up the product rule.\[\large y'=\color{royalblue}{(x^3)'}\sin^2(5x)+x^3\color{royalblue}{(\sin^2(5x))'}\]

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0I already got to that point @zepdrix that's where I am stuck :D 2x^2sin(5x)+2x^3sin(5x)

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0I forgot a power of 2 on the first sin *

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2\[\large y'=\color{orangered}{3x^2}\sin^2(5x)+x^3\color{royalblue}{(\sin^2(5x))'}\]I've taken the derivative of the first one. I think your coefficient is a lil mixed up. You brought the wrong exponent down :O

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0Bro which one are you referring to

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2The first blue term in the original product rule that I wrote out.

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0That? Am I not supposed to take down the power and multiply it with whatever it is next to it and and substract the power by 1?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2x^3. Bringing the power down and subtracting by one gives us, 3x^2 right? :o

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0I am talking about the power of sin not x

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0Remember now we differantiating the second part of the expression or whatever

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2I wasn't talking about that one. I was doing the other term.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2The second one will be a lil more complicated, I wanted to make sure you got the first part.

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0ah yea you got a point :D That one I didnt notice :D ok

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2So for the second one, again we'll apply the power rule to the sine function, then we'll apply the chain rule, multiplying by the derivative of the inside. \[\large y'=\color{orangered}{3x^2}\sin^2(5x)+x^3\color{orangered}{(2\sin(5x))}\color{royalblue}{(\sin(5x))'}\]The new blue term that showed up is the one we need to take a derivative of. If that is confusing, there is another way we can write it.

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0What we do this only for the second part if I may ask

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0Just need to understand whats the difference that forces us to do so

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2So we start with the product of two things involving x. \[\large y=(x^3)(\sin^2(5x))\] Without taking any derivatives yet, we can setup the product rule. \[\large y'=\color{royalblue}{(x^3)'}\sin^2(5x)+x^3\color{royalblue}{(\sin^2(5x))'}\] The blue terms are the ones we have to differentiate (take derivatives of). So you can see in our setup, we only differentiate the x^3 in the first term, and the sine in the second term. Was that the question? :o It's because of the product rule.

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0No I mean after that we do we take chain rule for Only the second term? Chain rule appears to be a lil messed up in my mind

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Yah the second blue term has an inner function. Umm I'm trying to think of a good way to explain it...

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0if I have this (y5x)^2 Will I take chain rule for derivative?

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0(5x)^2 *** ignore the fact that I can just do it 25x^2 lets say I COULDNT. Will I use chain rule?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Yes that's a very good example :) Here's how we would work it out.\[\large \left[(5x)^2\right]' \qquad = \qquad 2(5x)\]That's the derivative of the OUTER FUNCTION. (Something squared). Now we have to multiply by the derivative of the inside.\[\large \left[(5x)^2\right]' \qquad = \qquad 2(5x)(5x)' \qquad = \qquad 2(5x)(5)\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2The chain rule can be really tricky to get a handle on. Lemme know if you need another example.

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0I see :) Alright I got it. Just something last because according to what I learned now I still cant make the solution like that in the solution manual (unofficial manual) http://screencast.com/t/snYsnYAbha maybe the above link is wrong?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2No those steps are correct but they made a substitution to try and get the point across. I find that more confusing personally.. hmm

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0My own solution is 3x^2sin^2(5x)+2x^3sin(5x)cos(5x)

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0oh sh*t there is a 5 too hold on

Christos
 3 years ago
Best ResponseYou've already chosen the best response.03x^2sin^2(5x)+10x^3sin(5x)cos(5x)

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Yes very good :) So you applied the chain rule TWICE, yes? After you got a cosine, you took the derivative of the (5x)

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0but still the manual doesn't end up with any cos

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2What they did in the solutions manual is, they applied the `Double Angle Formula for Sine`. That's why it looks a bit different.\[\large 2\sin x \cos x = \sin 2x\]So if you look at your second term, ignoring the x^2 in front, we have something like this,\[\large 5\cdot2 \sin(5x)\cos(5x)\]Applying the double angle formula gives us,\[\large 5 \cdot \sin(10x)\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2They did have a cosine if you look at the middle steps. They applied an annoying trig rule to get rid of it though :)

Christos
 3 years ago
Best ResponseYou've already chosen the best response.0Not important for my course, im glad my way is correct too, thanks again
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