Differentiation Question

Differentiation sincubed2x

Right, so i've split that into sin2xsinx
and assigned

u = sin2x, du = 2cos2x
v = sinx, dv = cosx

then used product rule to find sin2xcosx + 2cos2xsinx

Is this correct? Answer says 6sinsquared2xcox2x

Reply 1

Smaug123

Original post by Axion

I assume you mean

\sin^3(2x)

; this is not the same as

\sin^2(x) \sin(x)

\sin(2x) \sin(x)

, so I think your first line is dodgy.

(edited 10 years ago)

Reply 2

Axion

Original post by Smaug123

I assume you mean

\sin^3(2x)

; this is not the same as

\sin^2(x) \sin(x)

\sin(2x) \sin(x)

, so I think your first line is dodgy.

Ah right.

Is it meant to be sin2x^2 * sin2x?

Reply 3

Smaug123

Original post by Axion

Ah right.

Is it meant to be sin2x^2 * sin2x?

Can you LaTeX it? I still can't really tell what you mean. I think you've written

\sin^2(2x) \sin(2x)

, which is indeed what it is.

Reply 4

Axion

Original post by Smaug123

Can you LaTeX it? I still can't really tell what you mean. I think you've written

\sin^2(2x) \sin(2x)

, which is indeed what it is.

Apologies, that is what I meant.

How would you differentiate sin^2(2x)?

2cos^2(2x)?

Reply 5

Smaug123

Original post by Axion

Apologies, that is what I meant.

How would you differentiate sin^2(2x)?

2cos^2(2x)?

How would you differentiate

f(2x)^2

? You're nearly right.

Reply 6

Axion

Original post by Smaug123

How would you differentiate

f(2x)^2

? You're nearly right.

oh right, its the same as

sin(2x)^2

So chain rule?

2*2cos2x = 4cos2x?

if that is the case it would be udv+vdu

= 2sinsq2xcosx + 4cos2xsin2x?

(edited 10 years ago)

Reply 7

Axion

My answer is definitely not right :/

Reply 8

davros

Original post by Axion

You can use the product rule on this but it will get messy!

This is just a basic application of the chain rule - you can write:

y = sin^3 u

with

u = 2x

and then just use

\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

Reply 9

Brit_Miller

Could just use the chain rule on the original

sin^3(2x)

Reply 10

Axion

how would you do it using the product rule though? Out of curiosity. Rep for the first answer :P

Reply 11

natninja

Original post by Axion

just use the chain rule, don't over-complicate stuff where you don't need to...

Reply 12

davros

Original post by Axion

how would you do it using the product rule though? Out of curiosity. Rep for the first answer :P

You'd need to write y = (sin 2x)(sin 2x)(sin 2x) i.e. uvw for 3 identical functions u,v and w and then differentiate that thing and put it back together again :smile:

Reply 13

Axion

Original post by davros

You'd need to write y = (sin 2x)(sin 2x)(sin 2x) i.e. uvw for 3 identical functions u,v and w and then differentiate that thing and put it back together again :smile:

oh god

Reply 14

Smaug123

Original post by Axion

oh god

It could be worse - at least here there's three-way symmetry going on! It would indeed be foul if it were

e^x \sin(x) \log(x)

…

Reply 15

Axion

Original post by Smaug123

It could be worse - at least here there's three-way symmetry going on! It would indeed be foul if it were

e^x \sin(x) \log(x)

…

Don't go there bro! rep for your help anwyay :smile:

Reply 16

Smaug123

Original post by Axion

Don't go there bro! rep for your help anwyay :smile:

Thanks

Reply 17

alo9952

Can't you just ---> 3sin^2(2x)*cos(2x)*2

Posted from TSR Mobile

Reply 18

Smaug123

Original post by alo9952

Can't you just ---> 3sin^2(2x)*cos(2x)*2

Wherever possible try to guide the asker to the answer rather than just giving the answer straight off! Not much is learned by doing this and the asker may not understand why the answer is what it is.

From http://www.thestudentroom.co.uk/wiki/Study_Help_Guidelines

Reply 19

alo9952

Yeah sorry I mean differentiate just as you would do with y=f(x)ⁿ
Like:

dy/dx = n*f(x)^n-1*f'(x)

It's much easier than trying to break sin(2x)² to smaller powers and then applying product rule.