Core 4 question find dy/dx

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
Enter our travel-writing competition for the chance to win a Nikon 1 J3 camera 21-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
Search Thread
Advanced Search
Sign in to Reply
  1. Nerdatious's Avatar
    1 19-05-2012  19:44
    • Respected Member
    • Posts: 189
    Core 4 question find dy/dx
    Given that y = 3^x , find dy/dx in terms of x

    Please help me! I know that you have to use ln but I am unsure as how to!

    Thanks !
  2. dean01234's Avatar
    2 19-05-2012  19:49
    • Exalted and Worshipped Member
    • Posts: 1,281
    Re: Core 4 question find dy/dx
    y=3^x

    lny = ln3^x

    lny=xln3

    x= lny / ln3

    dx= 1/ln3 * 1/y
    dy

    dx = 1/ yln3
    dy

    dy= ylna = (3^x) ln3
    dx

    whats with the neg?
    Last edited by dean01234; 25-05-2012 at 17:45.
    Post rating:
    1
    1
  3. Rational Paradox's Avatar
    3 19-05-2012  19:51
    • Adored and Respected Member
    • Location: United Kingdom
    • Posts: 540
    Re: Core 4 question find dy/dx
    (Original post by Nerdatious)
    Given that y = 3^x , find dy/dx in terms of x

    Please help me! I know that you have to use ln but I am unsure as how to!

    Thanks !
    Well you know ln(a^x) = xln(a), and should know how to differentiate e^bx for some constant b
  4. Nerdatious's Avatar
    4 19-05-2012  19:51
    • Respected Member
    • Posts: 189
    Re: Core 4 question find dy/dx
    Can you explain the 4th step please?
  5. Nerdatious's Avatar
    5 19-05-2012  19:53
    • Respected Member
    • Posts: 189
    Re: Core 4 question find dy/dx
    (Original post by dean01234)
    y=3^x

    lny = ln3^x

    lny=xln3

    x= lny / ln3

    dx = 1/ yln3
    dy

    dy= ylna = (3^x) ln3
    dx
    Can you explain the 4th step please?
  6. Implication's Avatar
    6 19-05-2012  19:54
    • Benevolent Member
    • Posts: 890
    Re: Core 4 question find dy/dx
    Isn't the derivative of ax a standard one we can just learn? You can do it the way dean01234 has described, or you can just remember that d/dx(ax)=axlna.

    Spoiler:
    Show
    Slightly shorter way:

    y=a^{x}
    ln(y)=ln(a^{x})=xln(a)
    Differentiating implicitly,
    \frac{1}{y} \frac{dy}{dx} = ln(a)
    \frac{dy}{dx}=yln(a)=a^{x}ln(a)
    Last edited by Implication; 19-05-2012 at 19:56.
  7. dean01234's Avatar
    7 19-05-2012  19:54
    • Exalted and Worshipped Member
    • Posts: 1,281
    Re: Core 4 question find dy/dx
    (Original post by Nerdatious)
    Can you explain the 4th step please?
    Sorry, I'm not entireley sure what happens there... but that is the proof for (y=a^x =dy/dx = a^x lna) that I have copied down


    I think it is just differentiation, but I haven't worked it out
    Last edited by dean01234; 19-05-2012 at 19:56.
  8. Nerdatious's Avatar
    8 19-05-2012  19:55
    • Respected Member
    • Posts: 189
    Re: Core 4 question find dy/dx
    (Original post by dean01234)
    Sorry, I'm not entireley sure what happens there... but that is the proof for (y=a^x =dy/dx = a^x lna) that I have copied down
    Okay thank you, I will just learn that then Thanks for your help.
  9. electriic_ink's Avatar
    9 19-05-2012  19:59
    • TSR Demigod
    • Location: London
    • Posts: 5,636
    Re: Core 4 question find dy/dx
    (Original post by Nerdatious)
    Okay thank you, I will just learn that then Thanks for your help.
    Or, if it helps, you could use that fact that 3^x = (e^{\ln 3})^x = e^{x \ln 3}.
  10. Nerdatious's Avatar
    10 19-05-2012  20:02
    • Respected Member
    • Posts: 189
    Re: Core 4 question find dy/dx
    (Original post by electriic_ink)
    Or, if it helps, you could use that fact that 3^x = (e^{\ln 3})^x = e^{x \ln 3}.
    Sorry I don't understand, how would I go from there?
  11. electriic_ink's Avatar
    11 19-05-2012  20:03
    • TSR Demigod
    • Location: London
    • Posts: 5,636
    Re: Core 4 question find dy/dx
    (Original post by Nerdatious)
    Sorry I don't understand, how would I go from there?
    d/dx (3^x)
    = d/dx (e^(x ln 3)) using the identity
    = e^(x ln 3) ln 3 since ln 3 is a constant.
    = 3^x ln 3 using the identity again
  12. Implication's Avatar
    12 19-05-2012  20:04
    • Benevolent Member
    • Posts: 890
    Re: Core 4 question find dy/dx
    (Original post by Nerdatious)
    Can you explain the 4th step please?
    x=\frac{ln(y)}{ln(a)}=\frac{1}{l n(a)} \times ln(y)
    \frac{dx}{dy}=\frac{1}{ln(a)} \times \frac{1}{y} = \frac{1}{yln(a)}
    \therefore \frac{dy}{dx}=yln(a)=a^{x}ln(a)

    Or just see the spoiler in my earlier post
    Last edited by Implication; 19-05-2012 at 20:08.
  13. owen1994's Avatar
    13 24-05-2012  13:29
    • Exalted Member
    • Posts: 379
    Re: Core 4 question find dy/dx
    (Original post by Implication)
    x=\frac{ln(y)}{ln(a)}=\frac{1}{l n(a)} \times ln(y)
    \frac{dx}{dy}=\frac{1}{ln(a)} \times \frac{1}{y} = \frac{1}{yln(a)}
    \therefore \frac{dy}{dx}=yln(a)=a^{x}ln(a)

    Or just see the spoiler in my earlier post
    Can you or any of the other people explain Question 4C in Edexcel Core 4, the newish book.

    Find dy/dx for each of the following:

    c.) y=xa^x - The CD answers doesn't show me how to do it at all and just tells me they have used the product rule


    d.) y = 2^x / x - Then the quotient rule with this one. The CD again is just terrible and I need to know how to do these.

    thanks
  14. Implication's Avatar
    14 25-05-2012  10:45
    • Benevolent Member
    • Posts: 890
    Re: Core 4 question find dy/dx
    (Original post by owen1994)
    c.) y=xa^x - The CD answers doesn't show me how to do it at all and just tells me they have used the product rule
    Product rule: \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}
    y=xa^{x}
    So \frac{dy}{dx}=x \times \frac{d}{dx}(a^{x}) + a^{x} \times \frac{d}{dx}(x)=xa^{x}ln(a) + a^{x} = a^{x}(xln(a)+1)


    (Original post by owen1994)
    d.) y = 2^x / x - Then the quotient rule with this one. The CD again is just terrible and I need to know how to do these.
    Product rule: \frac{d}{dx}(\frac{u}{v})=\dfrac {v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^{2}}
    y=\frac{2^{x}}{x}
    Sp \frac{dy}{dx}=\dfrac{x \times \frac{d}{dx}(2^{x}) - 2^{x} \times \frac{d}{dx}(x)}{x^{2}}=\dfrac{x \cdot 2^{x} \cdot ln(2) - 2^{x}}{x^{2}}=\dfrac{2^{x}(xln(2 )-1)}{x^{2}}


    Does that make any sense? I usually remember the two rules like this:
    Product rule - "the first times the derivative of the second plus the second times the derivative of the first"
    Quotient rule - "the bottom times the derivative of the top minus the top times the derivative of the bottom, all over the bottom squared"

    Also, isn't this C3? That may be why they haven't explained it in the C4 book
  15. owen1994's Avatar
    15 25-05-2012  16:12
    • Exalted Member
    • Posts: 379
    Re: Core 4 question find dy/dx
    (Original post by Implication)
    Product rule: \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}
    y=xa^{x}
    So \frac{dy}{dx}=x \times \frac{d}{dx}(a^{x}) + a^{x} \times \frac{d}{dx}(x)=xa^{x}ln(a) + a^{x} = a^{x}(xln(a)+1)




    Product rule: \frac{d}{dx}(\frac{u}{v})=\dfrac {v \times \frac{du}{dx} - u \times \frac{dv}{dx}}{v^{2}}
    y=\frac{2^{x}}{x}
    Sp \frac{dy}{dx}=\dfrac{x \times \frac{d}{dx}(2^{x}) - 2^{x} \times \frac{d}{dx}(x)}{x^{2}}=\dfrac{x \cdot 2^{x} \cdot ln(2) - 2^{x}}{x^{2}}=\dfrac{2^{x}(xln(2 )-1)}{x^{2}}


    Does that make any sense? I usually remember the two rules like this:
    Product rule - "the first times the derivative of the second plus the second times the derivative of the first"
    Quotient rule - "the bottom times the derivative of the top minus the top times the derivative of the bottom, all over the bottom squared"

    Also, isn't this C3? That may be why they haven't explained it in the C4 book
    I know all the rules from C3 and the rules for implicit equations I just can't get my head round these questions. I still don't really understand, but thanks for trying. I actually ended up just skipping them two questions and I moved on to finish the rest of differentiation. I only have Vectors and Integration to finish now, hopefully get vectors done this weekend . Thanks for help.
    Post rating:
    1
    1
  16. Thasneemy's Avatar
    16 25-05-2012  16:18
    • Full Member
    • Posts: 106
    Re: Core 4 question find dy/dx
    If y = a^x

    dy/dx = a^x ln a for any value of a.

    Someone correct me if i'm wrong please, i'm saying this off the top of my head, but i'm pretty sure this is the equation we need to know.

    Thanks for the negs, didnt know people hate maths that much.
    Last edited by Thasneemy; 23-06-2012 at 17:09.
    Post rating:
    1
    2
  17. electriic_ink's Avatar
    17 25-05-2012  16:26
    • TSR Demigod
    • Location: London
    • Posts: 5,636
    Re: Core 4 question find dy/dx
    (Original post by Thasneemy)
    If y = a^x

    dy/dx = a^x ln a for any value of a.

    Someone correct me if i'm wrong please, i'm saying this off the top of my head, but i'm pretty sure this is the equation we need to know.
    Why not read the thread and find out? :yy:
    Post rating:
    1
    2
  18. Thasneemy's Avatar
    18 25-05-2012  16:37
    • Full Member
    • Posts: 106
    Re: Core 4 question find dy/dx
    (Original post by owen1994)
    Can you or any of the other people explain Question 4C in Edexcel Core 4, the newish book.

    Find dy/dx for each of the following:

    c.) y=xa^x - The CD answers doesn't show me how to do it at all and just tells me they have used the product rule


    d.) y = 2^x / x - Then the quotient rule with this one. The CD again is just terrible and I need to know how to do these.

    thanks
    Hey!

    I dont know if you've solved these questions or whether you understand it now. But i'm going to try and help you through it. Forgive me if you understand the process, I can understand it may be annoying to be told the same thing over and over! Maybe by the time i've written this all out, someone's got there before me!


    Ok, first general rule to keep in mind: if y = a^x then dy/dx= a^x lna

    Question c: y=xa^x

    Here, you have two things multiplying each other. you have x being multiplied by a^x. So immediately, you realise the need for product rule.

    Product rule is: u(dv/dx) + v(du/dx)

    First, lets establish what v is, what dv/dx is, what u is, what du/dx is.

    v=x therefore dv/dx= 1

    u=a^x therefore du/dx= a^xlna (using the first equation I showed you: y = a^x then dy/dx= a^x lna)

    Ok, now, all you do is simply plug it into the product rule!

    (x) x (a^xlna) + (1) x (a^x) ---> xa^xlna + a^x

    Now, all you do is take out a common factor of a^x so you get:

    a^x(xlna + 1) and that's done!


    Question d: y = 2^x / x

    Notice two things are being divided, so you immediately recognise the need for the quotient rule.

    The quotient rule is: v(du/dx) - u(dv/dx) all divided by v^2.

    Again, let's establish v, dv/dx, u and du/dx.

    v= x therefore dv/dx= 1

    u= 2^x therefore du/dx= 2^xln2 (using the same initial formula for powers)

    Now, simply plug it all into the quotient rule!

    ((x) x (2^xln2) - (2^x)) / (x^2) (write this out on paper, it will look clearer.)

    Then, you may take out from that a common factor of 2^x, leaving you with:

    (2^x(xln2 - 1)) / x^2 and that's done!

    Hope it makes sense!
    Last edited by Thasneemy; 25-05-2012 at 16:40.
  19. Thasneemy's Avatar
    19 25-05-2012  16:38
    • Full Member
    • Posts: 106
    Re: Core 4 question find dy/dx
    (Original post by electriic_ink)
    Why not read the thread and find out? :yy:
    Thanks, i'll bear that in mind in the future.
  20. Implication's Avatar
    20 25-05-2012  18:08
    • Benevolent Member
    • Posts: 890
    Re: Core 4 question find dy/dx
    (Original post by owen1994)
    I know all the rules from C3 and the rules for implicit equations I just can't get my head round these questions. I still don't really understand, but thanks for trying. I actually ended up just skipping them two questions and I moved on to finish the rest of differentiation. I only have Vectors and Integration to finish now, hopefully get vectors done this weekend . Thanks for help.
    Fair enough. Maybe it's worth going over some more C3 differentiation questions, just so you get used to applying the rules (rather than just knowing them). Practice really does make perfect for a lot of things in maths

    I don't know why you were negged for that either There seems to be a lot of negging going on in this thread aha (please don't get me too ) aha. Anyway, good luck.
Sign in to Reply
Search Thread
Advanced Search
Share this discussion:  
Useful resources

Articles:

Study Help rules and posting guidelinesStudy Help Member of the Month nominationsGuide to MathematicsMathematics resourcesMathematics websitesMathematics revision notes in the TSR wikiCambridge Assessment admissions tests (including STEP)How to use LaTeXCareers in Mathematics

Useful Dates:

Quick Link:

Unanswered Maths Threads

Groups associated with this forum:

View associated groups
Article updates
Moderators

We have a brilliant team of more than 60 volunteers looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is moderated by:

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22

Registered Office: International House, Queens Road, Brighton, BN1 3XE

Shortcuts

Get Started

TSR Group

Using TSR

Info

Connect with TSR

Reputation gems:
The Reputation gems seen here indicate how well reputed the user is, red gem indicate negative reputation and green indicates a good rep.
Post rating score:
These scores show if a post has been positively or negatively rated by our members.