Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    16
    ReputationRep:
    Okay, the book says that if y=a^x, \text{then }\frac{dy}{dx}=a^xlna. Then, in one of the questions I need to differentiate 2^{-x}. So I use the general rule above and get \frac{dy}{dx}=2^{-x}ln2, however they got \frac{dy}{dx}=2^{-x}ln2. Am I missing something here?

    Second question:

    The book explains how to differentiate the general power function:
    \begin{array}{lcr}

y=a^x\\

lny=xlna\\

\therefore \frac{1}{y}\cdot \frac{dy}{dx}=lna

\end{array}

    Ok.. stopping there, surely that's not correct? They differentiated lny, and by using the chain rule got \frac{1}{y}\cdot \frac{dy}{dx}. Then, by definition, shouldn't you get lna+x\cdot \frac{1}{a}\cdot \frac{da}{dx} for the RHS? How did they just get lna?
    Offline

    17
    ReputationRep:
    (Original post by ViralRiver)
    Okay, the book says that if y=a^x, \text{then }\frac{dy}{dx}=a^xlna. Then, in one of the questions I need to differentiate 2^{-x}. So I use the general rule above and get \frac{dy}{dx}=2^{-x}ln2, however they got \frac{dy}{dx}=2^{-x}ln2. Am I missing something here?
    The book is wrong and so are you.
    Note that 2^{-x} = (\frac{1}{2})^x. Apply your rule to this and you will get \frac{dy}{dx} = -2^{-x}\ln 2.

    Second question:

    The book explains how to differentiate the general power function:
    \begin{array}{lcr}

y=a^x\\

lny=xlna\\

\therefore \frac{1}{y}\cdot \frac{dy}{dx}=lna

\end{array}

    Ok.. stopping there, surely that's not correct? They differentiated lny, and by using the chain rule got \frac{1}{y}\cdot \frac{dy}{dx}. Then, by definition, shouldn't you get lna+x\cdot \frac{1}{a}\cdot \frac{da}{dx} for the RHS? How did they just get lna?
    That's correct. When you differentiate nx w.r.t. x, you get n. (\ln a)x is of this form.
    Offline

    12
    ReputationRep:
    I'm guessing you mean \frac{dy}{dx}= -2^{-x} ln 2. Remember to use the chain rule.

    As for the second part, a is a constant, so its derivative is 0.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by Farhan.Hanif93)
    The book is wrong and so are you.
    Note that 2^{-x} = (\frac{1}{2})^x. Apply your rule to this and you will get \frac{dy}{dx} = -2^{-x}\ln 2.


    That's correct. When you differentiate nx w.r.t. x, you get n. (\ln a)x is of this form.
    SOrry, I meant the book got -2^-xln2, which is what I don't understand. What do you mean by your response to the second question?
    Offline

    0
    ReputationRep:
    similarly, what's \displaystyle\int ex\ dx ?
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by Pheylan)
    similarly, what's \displaystyle\int ex\ dx ?
    \frac{ex^2}{2}+c
    Offline

    17
    ReputationRep:
    (Original post by ViralRiver)
    SOrry, I meant the book got -2^-xln2, which is what I don't understand.
    \frac{d}{dx}[2^{-x}]
    = \frac{d}{dx}\left[(\frac{1}{2})^x\right]
    =(\frac{1}{2})^x \ln (\frac{1}{2})
    =2^{-x}\ln (2^{-1})
    =-2^{-x}\ln 2

    What do you mean by your response to the second question?
    a is not a variable, it's a constant. Therefore lna is a constant. When you differentiate nx for some constant coefficient, n, you get the derivative to be n, right? C1 differentiation?
    So \frac{d}{dx}[x\ln a] = \ln a.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by Pork and Beans)
    I'm guessing you mean \frac{dy}{dx}= -2^{-x} ln 2. Remember to use the chain rule.

    As for the second part, a is a constant, so its derivative is 0.
    Well, how do we know a is a constant? I'm still unsure of how to get -2 etc for the first part, even with the chain rule >< - this doesn't seem to make any sense.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by Farhan.Hanif93)
    \frac{d}{dx}[2^{-x}]
    = \frac{d}{dx}\left[(\frac{1}{2})^x\right]
    =(\frac{1}{2})^x \ln (\frac{1}{2})
    =2^{-x}\ln (2^{-1})
    =-2^{-x}\ln 2


    a is not a variable, it's a constant. Therefore lna is a constant. When you differentiate nx for some constant coefficient, n, you get the derivative to be n, right? C1 differentiation?
    So \frac{d}{dx}[x\ln a] = \ln a.
    Okay, I understand it a bit more now, but I still have a few problems. If y = 2^{-x} then y = \frac{1}{2^x}, which doesn't make a difference in this question, but why did you raise the entire fraction (1/2) to the power of x?
    Offline

    17
    ReputationRep:
    (Original post by ViralRiver)
    Okay, I understand it a bit more now, but I still have a few problems. If y = 2^{-x} then y = \frac{1}{2^x}, which doesn't make a difference in this question, but why did you raise the entire fraction (1/2) to the power of x?
    To apply the rule you stated. You can always do it the long way. To be in the form a^x, which you want, you need the coefficient of x in the power to be 1. It's still equivalent because 1^x = 1.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by Farhan.Hanif93)
    To apply the rule you stated. You can always do it the long way. To be in the form a^x, which you want, you need the coefficient of x in the power to be 1. It's still equivalent because 1^x = 1.
    Perfect, thanks .

    Is there a sort of general rule in a similar way to the following:?

    \frac{d(sinf(x))}{dx}=f'(x)cos(f  (x))
    Offline

    17
    ReputationRep:
    (Original post by ViralRiver)
    Perfect, thanks .

    Is there a sort of general rule in a similar way to the following:?

    \frac{d(sinf(x))}{dx}=f'(x)cos(f  (x))
    Yes, that is true.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by Farhan.Hanif93)
    Yes, that is true.
    Hmm, soryr I should be more clear.

    I can use the chain rule to differentiate sin, as follows:

    f(x)=h(g(x))
    f'(x)=g'(x)\cdot h'(g(x)

    then, if

    f(x) = sin(x^2), f'(x) = 2xcos(x^2)

    but.. what if f(x) = a^x, how do you use the chain rule on that?
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.