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# Core 4 Differentiation

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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
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because

ln(sinx)

is ln of sinx

not ln time sinx
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All of the above use the Chain Rule
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(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
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(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
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(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 gives

If , then
Hence,
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(Original post by raheem94)
1) e.g. differentiate ln(cosx)
We know differentiating gives

If , then
Hence,
I though that if you differentiate lnx you get 1/x or is that integrating it? :/
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(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?
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(Original post by fishfingers:))
Whats the differential of lnx?
1/x

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

So in the case of ,
differentiating f(x) gives hence
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(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
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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.
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(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 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?
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(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!
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(Original post by raheem94)
Differentiating the function gives

So in the case of ,
differentiating f(x) gives hence
thanks ! how do i work out question 2 ??
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(Original post by fishfingers:))
thanks ! how do i work out question 2 ??
Remember differentiating gives,

Can you do it now?
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(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.
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(Original post by raheem94)
Remember differentiating gives,

Can you do it now?

Yep thank Yoouuu x
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(Original post by fishfingers:))
Yep thank Yoouuu x
No problem, you are welcome.
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(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:

1:

2:

3:

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!

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