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

1. abb0t

product rule within chain rule.

2. zepdrix

$\large y=x^3 \sin^2(5x)$We start by setting up the product rule.$\large y'=\color{royalblue}{(x^3)'}\sin^2(5x)+x^3\color{royalblue}{(\sin^2(5x))'}$

3. Christos

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

4. Christos

I forgot a power of 2 on the first sin *

5. 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

6. Christos

Bro which one are you referring to

7. zepdrix

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

8. 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?

9. zepdrix

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

10. Christos

I am talking about the power of sin not x

11. Christos

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

12. zepdrix

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

13. zepdrix

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

14. Christos

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

15. 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.

16. Christos

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

17. Christos

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

18. Christos

why*

19. 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.

20. 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

21. zepdrix

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

22. Christos

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

23. Christos

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

24. 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.$\large \left[(5x)^2\right]' \qquad = \qquad 2(5x)(5x)' \qquad = \qquad 2(5x)(5)$

25. zepdrix

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

26. 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?

27. 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

28. Christos

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

29. Christos

oh sh*t there is a 5 too hold on

30. Christos

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

31. zepdrix

Yes very good :) So you applied the chain rule TWICE, yes? After you got a cosine, you took the derivative of the (5x)

32. Christos

yea

33. Christos

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

34. 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,$\large 5 \cdot \sin(10x)$

35. 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 :)

36. Christos

ooh so thats why

37. Christos

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

38. zepdrix

np c: