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Core 4 question find dy/dx

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    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 !
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    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?
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    (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
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    Can you explain the 4th step please?
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    (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?
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    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)
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    (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
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    (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.
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    (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}.
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    (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?
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    (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
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    (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
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    (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
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    (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
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    (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.
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    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.
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    (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:
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    (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!
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    (Original post by electriic_ink)
    Why not read the thread and find out? :yy:
    Thanks, i'll bear that in mind in the future.
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    (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.

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