Find derivative of http://screencast.com/t/sXZqTgJKM
Just tell me 1-2 steps I will clip to them afterwards

- Christos

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- schrodinger

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- abb0t

product rule within chain rule.

- 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

I already got to that point @zepdrix that's where I am stuck :D
2x^2sin(5x)+2x^3sin(5x)

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## More answers

- Christos

I forgot a power of 2 on the first sin *

- 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

Bro which one are you referring to

- zepdrix

The first blue term in the original product rule that I wrote out.

- 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

x^3.
Bringing the power down and subtracting by one gives us,
3x^2
right? :o

- Christos

I am talking about the power of sin not x

- Christos

Remember now we differantiating the second part of the expression or whatever

- zepdrix

I wasn't talking about that one.
I was doing the other term.

- zepdrix

The second one will be a lil more complicated, I wanted to make sure you got the first part.

- Christos

ah yea you got a point :D That one I didnt notice :D ok

- 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

What we do this only for the second part if I may ask

- Christos

Just need to understand whats the difference that forces us to do so

- Christos

why*

- 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

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

Yah the second blue term has an inner function.
Umm I'm trying to think of a good way to explain it...

- Christos

if I have this (y5x)^2
Will I take chain rule for derivative?

- Christos

(5x)^2 *** ignore the fact that I can just do it 25x^2 lets say I COULDNT. Will I use chain rule?

- 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

The chain rule can be really tricky to get a handle on.
Lemme know if you need another example.

- 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

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

My own solution is 3x^2sin^2(5x)+2x^3sin(5x)cos(5x)

- Christos

oh sh*t there is a 5 too hold on

- Christos

3x^2sin^2(5x)+10x^3sin(5x)cos(5x)

- 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

yea

- Christos

but still the manual doesn't end up with any cos

- 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

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

ooh so thats why

- Christos

Not important for my course, im glad my way is correct too, thanks again

- zepdrix

np c:

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