Christos
  • Christos
Find derivative of http://screencast.com/t/sXZqTgJKM Just tell me 1-2 steps I will clip to them afterwards
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
abb0t
  • abb0t
product rule within chain rule.
zepdrix
  • zepdrix
\[\large y=x^3 \sin^2(5x)\]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
  • Christos
I already got to that point @zepdrix that's where I am stuck :D 2x^2sin(5x)+2x^3sin(5x)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Christos
  • Christos
I forgot a power of 2 on the first sin *
zepdrix
  • zepdrix
\[\large y'=\color{orangered}{3x^2}\sin^2(5x)+x^3\color{royalblue}{(\sin^2(5x))'}\]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
  • Christos
Bro which one are you referring to
zepdrix
  • zepdrix
The first blue term in the original product rule that I wrote out.
Christos
  • Christos
That? 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
  • zepdrix
x^3. Bringing the power down and subtracting by one gives us, 3x^2 right? :o
Christos
  • Christos
I am talking about the power of sin not x
Christos
  • Christos
Remember now we differantiating the second part of the expression or whatever
zepdrix
  • zepdrix
I wasn't talking about that one. I was doing the other term.
zepdrix
  • zepdrix
The second one will be a lil more complicated, I wanted to make sure you got the first part.
Christos
  • Christos
ah yea you got a point :D That one I didnt notice :D ok
zepdrix
  • zepdrix
So 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
  • Christos
What we do this only for the second part if I may ask
Christos
  • Christos
Just need to understand whats the difference that forces us to do so
Christos
  • Christos
why*
zepdrix
  • zepdrix
So 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
  • Christos
No 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
  • zepdrix
Yah the second blue term has an inner function. Umm I'm trying to think of a good way to explain it...
Christos
  • Christos
if I have this (y5x)^2 Will I take chain rule for derivative?
Christos
  • Christos
(5x)^2 *** ignore the fact that I can just do it 25x^2 lets say I COULDNT. Will I use chain rule?
zepdrix
  • zepdrix
Yes 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
  • zepdrix
The chain rule can be really tricky to get a handle on. Lemme know if you need another example.
Christos
  • Christos
I 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
  • zepdrix
No those steps are correct but they made a substitution to try and get the point across. I find that more confusing personally.. hmm
Christos
  • Christos
My own solution is 3x^2sin^2(5x)+2x^3sin(5x)cos(5x)
Christos
  • Christos
oh sh*t there is a 5 too hold on
Christos
  • Christos
3x^2sin^2(5x)+10x^3sin(5x)cos(5x)
zepdrix
  • zepdrix
Yes very good :) So you applied the chain rule TWICE, yes? After you got a cosine, you took the derivative of the (5x)
Christos
  • Christos
yea
Christos
  • Christos
but still the manual doesn't end up with any cos
zepdrix
  • zepdrix
What 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
  • zepdrix
They did have a cosine if you look at the middle steps. They applied an annoying trig rule to get rid of it though :)
Christos
  • Christos
ooh so thats why
Christos
  • Christos
Not important for my course, im glad my way is correct too, thanks again
zepdrix
  • zepdrix
np c:

Looking for something else?

Not the answer you are looking for? Search for more explanations.