The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

C4 Maths - Differentiation... Help???

Scroll to see replies

Original post by sabahshahed294
First step(taking ln on both sides) is correct. Second part is also correct that you differentiate in terms of x to get 1y×dydx\frac{1}{y}\times\frac{dy}{dx}. And yes, use the property of logs so as to get xln3 x ln 3. You do this before differentiating with respect to x and after taking ln on both sides. Do you know what to do after that?


Not really, no. I'm struggling with how to differentiate the xln3 x ln 3 ... I end up with x3 \frac{x}{3} ... after differentiating this term.

And then I don't know where to go from there :frown:
(edited 7 years ago)
Original post by Philip-flop
Not really, no. I'm struggling with how to differentiate the xln3 x ln 3 ... I end up with x3 \frac{x}{3} ... after differentiating


Ln(3) is just a constant. Treat it the same way as if it would be 2x instead of xln(3)
ln3 is just a number ( slightly over 1 ) ...
Original post by RDKGames
Ln(3) is just a constant. Treat it the same way as if it would be 2x instead of xln(3)


Oh yeah of course it's just a constant. But I'm still being stupid as I don't know what to do.

So I could technically see... xln3x ln 3 ...as... x(1.098612289)or1.098612289x x(1.098612289) or 1.098612289x

So then... d(1.098612289x)dx=1.098612289 \frac{d(1.098612289x)}{dx} = 1.098612289 ???

But since.... 1.098612289=ln3 1.098612289 = ln 3

I can just say that d(xln3)dx=ln3 \frac{d(xln3)}{dx} = ln 3 ... correct?
(edited 7 years ago)
Original post by Philip-flop
Oh yeah of course it's just a constant. But I'm still being stupid as I don't know what to do.

So I could technically see... xln3x ln 3 ...as... x(1.098612289) x(1.098612289)

So then... d[x(1.098612289)]dx=1.098612289 \frac{d[x(1.098612289)]}{dx} = 1.098612289 ???

But since.... 1.098612289=ln3 1.098612289 = ln 3

I can just say that d(xln3)dx=ln3 \frac{d(xln3)}{dx} = ln 3 ... correct?


Yep
Original post by RDKGames
Yep


Thank you so much @RDKGames, I'm surprised you're not fed up with me by now!! :colondollar:

Original post by the bear
ln3 is just a number ( slightly over 1 ) ...


And thank you too :smile: :smile: :smile:
Am I writing too much for my workings for part (c)???...

C4 EXE4C Q1.png

Attachment not found
Photo 24-12-2016, 15 37 59.jpg156.3KB
Original post by Philip-flop
Am I writing too much for my workings for part (c)???...

C4 EXE4C Q1.png



yeah because you are not noticing that you can simply use the product rule here straight away.
(edited 7 years ago)
Original post by RDKGames
yeah because you are not noticing that you can simply use the product rule here straight away.


Straight away as in, use the product rule on y=xax y = xa^x ??

I was just going by what the book taught me :frown:
Original post by Philip-flop
Straight away as in, use the product rule on y=xax y = xa^x ??

I was just going by what the book taught me :frown:


Yes.
v=x and u=a^x
Original post by RDKGames
Yes.
v=x and u=a^x


Which implies that... dvdx=1\frac{dv}{dx} = 1 and dudx=axlna \frac{du}{dx} = a^x ln a ??? I have a feeling I differentiated the second part wrong :frown:
Original post by Philip-flop
Which implies that... dvdx=1\frac{dv}{dx} = 1 and dudx=axlna \frac{du}{dx} = a^x ln a ??? I have a feeling I differentiated the second part wrong :frown:


It's correct
Original post by RDKGames
It's correct


The annoying thing is I originally thought about using the Product Rule from the beginning but I decided to follow the same layout as example 6 from the textbook as I thought this was the only way :frown:

Thanks for pointing it out thought :smile: :smile: :smile:
Reply 33
Avatar for Notnek
Notnek
21
Original post by Philip-flop
The annoying thing is I originally thought about using the Product Rule from the beginning but I decided to follow the same layout as example 6 from the textbook as I thought this was the only way :frown:

Thanks for pointing it out thought :smile: :smile: :smile:

In C4 you can use the result that the derivative of axa^x is ln⁡a ax\ln a \ a^x without proof. Similar to how you use the fact that the derivative of sin⁡x\sin x is cos⁡x\cos x.

But you do need to know how to prove it using implicit differentiation and you may need to do this in the exam.
Original post by notnek
In C4 you can use the result that the derivative of axa^x is ln⁡a ax\ln a \ a^x without proof. Similar to how you use the fact that the derivative of sin⁡x\sin x is cos⁡x\cos x.

But you do need to know how to prove it using implicit differentiation and you may need to do this in the exam.

Thank yoooooou!

I've made note of how to prove it for future references.

How do you know all this @notnek ?? You actually know the layout of the Edexcel Maths exams inside out!!
Reply 35
Avatar for Naruke
Naruke
11
Original post by Philip-flop
Thank yoooooou!

I've made note of how to prove it for future references.

How do you know all this @notnek ?? You actually know the layout of the Edexcel Maths exams inside out!!


She used to be an exams officer
Merry Christmas!!!

Shout out to everyone who has helped me survive during my self-taught studies, especially @notnek and @RDKGames!! :smile:
Original post by Philip-flop
Merry Christmas!!!

Shout out to everyone who has helped me survive during my self-taught studies, especially @notnek and @RDKGames!! :smile:


Any time buddy :smile: MC!
Reply 38
Avatar for Zacken
Zacken
22
Original post by Naruke
She used to be an exams officer


She?
Also thanks @Zacken hope you're having a good Christmas too!

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you