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# Differentiating 3sin^2 x watch

1. why is dy/dx of 3sin^2 x = 6sin(x)cos(x)
Rather than 3cos^2 x
2. (Original post by chanel_666)
why is dy/dx of 3sin^2 x = 6sin(x)cos(x)
Rather than 3cos^2 x
It's the chain rule. Do you know how to differentiate something like

?

Your one is similar since you can write it as

If you can't do the non-trig one above then you won't be able to do the trig one.
3. (Original post by chanel_666)
why is dy/dx of 3sin^2 x = 6sin(x)cos(x)
Rather than 3cos^2 x

Using the chain rule the differential of 3(sinx)^2 is:

The differential of 3u^2 multiplied by the differential of u which in this case = sinx

So that would be 6u * cosx

And since u = sinx that makes it 6sinxcosx
4. (Original post by Notnek)
It's the chain rule. Do you know how to differentiate something like

?

Your one is similar since you can write it as

If you can't do the non-trig one above then you won't be able to do the trig one.
I can do the first one, I didn't think to use the chain rule for the trig one but I see why it works now thanks
5. why are you doing this at AS it's C3
6. (Original post by EmilySarah00)
Using the chain rule the differential of 3(sinx)^2 is:

The differential of 3u^2 multiplied by the differential of u which in this case = sinx

So that would be 6u * cosx

And since u = sinx that makes it 6sinxcosx
Thanks this makes sense

(Original post by _princessxox)
why are you doing this at AS it's C3
It's just a bit of cheeky self teach uno
7. (Original post by chanel_666)
It's just a bit of cheeky self teach uno
You’re welcome

Also nice one self teaching yourself this if you’re only in AS year, just make sure you prioritise your current subject content!
8. basically a quick way of doing it is by rewriting it as 3(sinx)^2 which is in the form [f(X)]^n.
It looks scary but stay with me on this as it will save you tons of time in answering these types of questions.
dy/dx would = n[f(x)]^(n-1) x f'(x) so in this example...

1) bring the n down to the front so it's 6 (2x3), 3 was just a constant.

2) write [f(x)] normally but to the power of (n-1) not n, so its 6[sin(x)]^1 or just 6[sin(x)]

3) now take whatever was in the square brackets [f(x)] i.e sin(x) and differentiate this and put it on the end so its 6[sin(x)] multiplied by cos(x) which is just 6sin(x)cos(x).

They key was rewriting sin^2(x) as [sin(x)]^2 remember this is squaring the whole bracket NOT (x)!!
9. (Original post by EmilySarah00)
You’re welcome

Also nice one self teaching yourself this if you’re only in AS year, just make sure you prioritise your current subject content!
LOL Ive finished my spec for AS I started in summer so Ive been revising
How would I differentiate 2e^x is it still the chain rule cos I didn't think it was needed
10. (Original post by chanel_666)
P

LOL Ive finished my spec for AS I started in summer so Ive been revising
How would I differentiate 2e^x is it still the chain rule cos I didn't think it was needed
The chain rule is used when you have a function within a function so e.g. , this is the function inside the function .

is more simple : it's just a constant multiplied by a standard function. When there's a constant you leave it alone. Does that help and do you know the answer now?
11. (Original post by chanel_666)
P

LOL Ive finished my spec for AS I started in summer so Ive been revising
How would I differentiate 2e^x is it still the chain rule cos I didn't think it was needed
Derivative of is an extremely standard thing to know at C3. Look it up and you'll never forget it, then your question becomes extremely trivial.
12. (Original post by RDKGames)
Derivative of is an extremely standard thing to know at C3. Look it up and you'll never forget it, then your question becomes extremely trivial.
It sounds like the OP is doing AS and self-teaching this stuff.
13. (Original post by Notnek)
The chain rule is used when you have a function within a function so e.g. , this is the function inside the function .

is more simple : it's just a constant multiplied by a standard function. When there's a constant you leave it alone. Does that help and do you know the answer now?
This makes a lot more sense thank u !!! Think I just need practise with differentiating trig
14. first apply power rule
2*3Sin(x)
then,
differentiate sin(x) which is cos(x) and multiply with 6Sin(x)
I.e.6Sin(x)Cos(x)

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