The Student Room Group

Core 4 Differentiation

Hey all!
I'm stuck on differentiating these questions...

1) ln(sinx) << why can't I just use the product rule on this? Like have u=ln and v=sinx and then differentiate the two to get du/dx = 1/x and dv/dx = cosx therefore giving sinx/x + lncosx? However, in the mark scheme it says...
1/sinx x cosx = cotx... But i don't get where they've got this from?

2)6e^tanx... I really don't know how to approach this at all
MARK SCHEME = 6e^tanx x sec^2x = 6e^(tanx)sec^2x

3) square root of cos2x MARK SCHEME = 1/2(cos2x)^-1/2 x (-2sin2x)
=-(sin2x)/square root of cos2x

If there is any kind person out there who could explain to me how the answer has been derived please, then I would greatly appreciate it
(edited 11 years ago)

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Reply 1
because

ln(sinx)

is ln of sinx

not ln time sinx
Reply 2
All of the above use the Chain Rule
Reply 3
Original post by TenOfThem
All of the above use the Chain Rule


so whats the formula for the chain rule? Because I usually just work it out without the formula although its difficult to do that with the above questions
Reply 4
Original post by fishfingers:)
so whats the formula for the chain rule? Because I usually just work it out without the formula although its difficult to do that with the above questions


I am not sure why these would be harder

You know the differential of ln and the differential of sin and you times them

ln(sinx)

diff the ln gives 1/sinx

diff the sinx gives cosx

times them

1/sinx times cosx = cosx/sinx = cotx
Reply 5
Original post by fishfingers:)
Hey all!
I'm stuck on differentiating these questions...

1) ln(sinx) << why can't I just use the product rule on this? Like have u=ln and v=sinx and then differentiate the two to get du/dx = 1/x and dv/dx = cosx therefore giving sinx/x + lncosx? However, in the mark scheme it says...
1/sinx x cosx = cotx... But i don't get where they've got this from?



1) e.g. differentiate ln(cosx)
We know differentiating ln(f(x)) \displaystyle ln(f(x)) gives f(x)f(x) \displaystyle \frac{f'(x)}{f(x)}

If f(x)=cosx \displaystyle f(x)=cosx, then f(x)=sinx \displaystyle f'(x)=-sinx
Hence, ddx(ln(cosx))=sinxcosx=tanx \displaystyle \frac{d}{dx}(ln(cosx)) = \frac{-sinx}{cosx}=-tanx
Reply 6
Original post by raheem94
1) e.g. differentiate ln(cosx)
We know differentiating ln(f(x)) \displaystyle ln(f(x)) gives f(x)f(x) \displaystyle \frac{f'(x)}{f(x)}

If f(x)=cosx \displaystyle f(x)=cosx, then f(x)=sinx \displaystyle f'(x)=-sinx
Hence, ddx(ln(cosx))=sinxcosx=tanx \displaystyle \frac{d}{dx}(ln(cosx)) = \frac{-sinx}{cosx}=-tanx


I though that if you differentiate lnx you get 1/x or is that integrating it? :/
Reply 7
Original post by TenOfThem
I am not sure why these would be harder

You know the differential of ln and the differential of sin and you times them

ln(sinx)

diff the ln gives 1/sinx

diff the sinx gives cosx

times them

1/sinx times cosx = cosx/sinx = cotx


Whats the differential of lnx?
Reply 8
Original post by fishfingers:)
Whats the differential of lnx?


1/x

which is where I got the 1/sinx from
Reply 9
Original post by fishfingers:)
I though that if you differentiate lnx you get 1/x or is that integrating it? :/


Differentiating the function ln(f(x)) \displaystyle ln(f(x)) gives f(x)f(x) \displaystyle \frac{f'(x)}{f(x)}

So in the case of lnx \displaystyle lnx ,
f(x)=x \displaystyle f(x) = x differentiating f(x) gives 1 \displaystyle 1 hence f(x)f(x)=1x \displaystyle \frac{f'(x)}{f(x)} = \frac{1}{x}
Reply 10
Original post by fishfingers:)
Hey all!
I'm stuck on differentiating these questions...

1) ln(sinx) << why can't I just use the product rule on this? Like have u=ln and v=sinx and then differentiate the two to get du/dx = 1/x and dv/dx = cosx therefore giving sinx/x + lncosx? However, in the mark scheme it says...
1/sinx x cosx = cotx... But i don't get where they've got this from?

2)6e^tanx... I really don't know how to approach this at all
MARK SCHEME = 6e^tanx x sec^2x = 6e^(tanx)sec^2x

3) square root of cos2x MARK SCHEME = 1/2(cos2x)^-1/2 x (-2sin2x)
=-(sin2x)/square root of cos2x

If there is any kind person out there who could explain to me how the answer has been derived please, then I would greatly appreciate it


my answer to your 1 st question is that ln 1 is = zero, so you cant use product rule with (ln1) x (sin x) as it would just give you the wrong answer.


havent bothered reading the rest of ure questions
Reply 11
oh and dont ask people on tsr for help as the majority of people who reply to ure thread are probably doing a level maths so they are not experts in maths and may not necessarily be giving you the correct advice, instead you should be asking your maths teacher.
Reply 12
Original post by TenOfThem
I am not sure why these would be harder

You know the differential of ln and the differential of sin and you times them

ln(sinx)

diff the ln gives 1/sinx

diff the sinx gives cosx

times them

1/sinx times cosx = cosx/sinx = cotx


thank you :biggrin: but i was just wondering how did you know that it was the chain rule? because i though that to use the chain rule there must be powers involved..

Also, how would i work out the second question then? becuase how do you differentiate e^tanx?? Is it just sec^2xe^tanx? :confused:
Reply 13
Original post by brownieboy
oh and dont ask people on tsr for help as the majority of people who reply to ure thread are probably doing a level maths so they are not experts in maths and may not necessarily be giving you the correct advice, instead you should be asking your maths teacher.


this is an a-level maths question lov!
Reply 14
Original post by raheem94
Differentiating the function ln(f(x)) \displaystyle ln(f(x)) gives f(x)f(x) \displaystyle \frac{f'(x)}{f(x)}

So in the case of lnx \displaystyle lnx ,
f(x)=x \displaystyle f(x) = x differentiating f(x) gives 1 \displaystyle 1 hence f(x)f(x)=1x \displaystyle \frac{f'(x)}{f(x)} = \frac{1}{x}


thanks ! how do i work out question 2 ??:tongue:
Reply 15
Original post by fishfingers:)
thanks ! how do i work out question 2 ??:tongue:


Remember differentiating ef(x) \displaystyle e^{f(x)} gives, f(x)ef(x) \displaystyle f'(x)e^{f(x)}

In your question f(x)=tanx \displaystyle f(x)=tanx

Can you do it now?
Reply 16
Original post by brownieboy
oh and dont ask people on tsr for help as the majority of people who reply to ure thread are probably doing a level maths so they are not experts in maths and may not necessarily be giving you the correct advice, instead you should be asking your maths teacher.


I am doing A-Level further maths, i already have an A* in A-Level maths, TenOfThem is a teacher, so who are you referring to people who won't be giving correct advice and are not experts in maths?

We don't have so much spare time to give people wrong advice.
(edited 11 years ago)
Reply 17
Original post by raheem94
Remember differentiating ef(x) \displaystyle e^{f(x)} gives, f(x)ef(x) \displaystyle f'(x)e^{f(x)}

In your question f(x)=tanx \displaystyle f(x)=tanx

Can you do it now?



Yep thank Yoouuu x
Reply 18
Original post by fishfingers:)
Yep thank Yoouuu x


No problem, you are welcome.
Original post by fishfingers:)
Hey all!
I'm stuck on differentiating these questions...

1) ln(sinx) << why can't I just use the product rule on this? Like have u=ln and v=sinx and then differentiate the two to get du/dx = 1/x and dv/dx = cosx therefore giving sinx/x + lncosx? However, in the mark scheme it says...
1/sinx x cosx = cotx... But i don't get where they've got this from?

2)6e^tanx... I really don't know how to approach this at all
MARK SCHEME = 6e^tanx x sec^2x = 6e^(tanx)sec^2x

3) square root of cos2x MARK SCHEME = 1/2(cos2x)^-1/2 x (-2sin2x)
=-(sin2x)/square root of cos2x

If there is any kind person out there who could explain to me how the answer has been derived please, then I would greatly appreciate it

The chain rule:
dydx=dydu×dudx\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}

1: y=lnu, u=sinxy=\ln u ,\ u=\sin x

2: y=6eu, u=tanxy=6e^u ,\ u=\tan x

3: y=u12, u=cos2xy=u^{\frac{1}{2}} ,\ u=\cos 2x

Whenever you have a function of a function like these examples, you can use the same technique. It can help avoid making careless mistakes if nothing else! :smile:

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