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Differentiating lnx watch

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    hi, need some help with this question
    find the equation of the normal of ln(x^2+1) where x = 1

    for dy/dx, i got 2x/(x^2+1) = 0, and I got 1 for the gradient, -1 for the gradient of the normal, im trying to find the y coordinate but im getting a faction when I put x into the original equation, am i doing something wrong?

    also find the coordinates of the point where the gradient of y = xlnx is 0,
    i got 1 + lnx for dy/dx, = 0, but how do i find the x value?

    thanks
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    (Original post by jacksonmeg)
    hi, need some help with this question
    find the equation of the normal of ln(x^2+1) where x = 1

    for dy/dx, i got 2x/(x^2+1) = 0, and I got 1 for the gradient, -1 for the gradient of the normal, im trying to find the y coordinate but im getting a faction when I put x into the original equation, am i doing something wrong?

    also find the coordinates of the point where the gradient of y = xlnx is 0,
    i got 1 + lnx for dy/dx, = 0, but how do i find the x value?

    thanks
    You have a function y=\ln(x^{2}+1)

    You are correct \dfrac{dy}{dx}=\dfrac{2x}{x^{2}+  1} It is equal to dy/dx not 0

    You are right -1 is the gradient of the normal so you now need to find the value of y when x=1

    When x=1 y=\ln(1^{2}+1)=... (it is not a fraction, it is a log you will end up with)
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    (Original post by Robbie242)
    You have a function y=\ln(x^{2}+1)

    You are correct \dfrac{dy}{dx}=\dfrac{2x}{x^{2}+  1} It is equal to dy/dx not 0

    You are right -1 is the gradient of the normal so you now need to find the value of y when x=1

    When x=1 y=\ln(1^{2}+1)=... (it is not a fraction, it is a log you will end up with)
    ln(2) = decimal ... idk what im doing wrong
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    (Original post by jacksonmeg)
    ln(2) = decimal ... idk what im doing wrong
    That's right when x=1 y=\ln(2) that's it

    You have a gradient now m=-1 and a point (1,ln2)

    Now you can find the equation of the normal by substituting this into a straight line equation
    y-y_{1}=m(x-x_{1})
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    (Original post by jacksonmeg)
    ln(2) = decimal ... idk what im doing wrong
    With C3, you need to learn to keep things like ln as they are. ln2 is ln2, use the formula y - y1 = m(x - x1).
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    (Original post by Robbie242)
    That's right when x=1 y=\ln(2) that's it

    You have a gradient now m=-1 and a point (1,ln2)

    Now you can find the equation of the normal by substituting this into a straight line equation
    y-y_{1}=m(x-x_{1})
    alright I got y = -x+ln2+1 which i checked in the back and is correct
    could you explain how to do the second one please?
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    (Original post by TSR561)
    With C3, you need to learn to keep things like ln as they are. ln2 is ln2, use the formula y - y1 = m(x - x1).
    I know getting a lot of e^x and lnx questions wrong by substituting values in and getting decimals ...
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    (Original post by jacksonmeg)
    alright I got y = -x+ln2+1 which i checked in the back and is correct
    could you explain how to do the second one please?
    Sure you need to find the value of x and the corresponding value of y such that \dfrac{dy}{dx}=0

    Well we know that \dfrac{dy}{dx}=1+\ln(x)

    We need to know when the gradient is zero so \dfrac{dy}{dx}=0

    Means \ln(x)=-1

    Now you need to remember that ln is the inverse of e and vice versa

    e^{\ln(f(x))}=f(x)

    So x=...

    then once you have x, plug it back into the equation for y which is y=x\ln(x)
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    (Original post by jacksonmeg)
    I know getting a lot of e^x and lnx questions wrong by substituting values in and getting decimals ...
    You'll get used to it haha. It's especially fun when you're integrating and can't use the integrating bit of your calculator (if you have one) to just give you the answer
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    (Original post by Robbie242)
    Sure you need to find the value of x and the corresponding value of y such that \dfrac{dy}{dx}=0

    Well we know that \dfrac{dy}{dx}=1+\ln(x)

    We need to know when the gradient is zero so \dfrac{dy}{dx}=0

    Means \ln(x)=-1

    Now you need to remember that ln is the inverse of e and vice versa

    e^{\ln(f(x))}=f(x)

    So x=...

    then once you have x, plug it back into the equation for y which is y=x\ln(x)
    you've lost me at the e^ln bit :/
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    (Original post by TSR561)
    You'll get used to it haha. It's especially fun when you're integrating and can't use the integrating bit of your calculator (if you have one) to just give you the answer
    we started integrating sin, cos, tan, ln, e^x and area under a curve last week, I prefer this to differentiation so far >:/
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    (Original post by jacksonmeg)
    you've lost me at the e^ln bit :/
    I think he's basically saying ln(x) is the same as loge(x)... so if loge(x) = -1, x = ...
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    (Original post by jacksonmeg)
    you've lost me at the e^ln bit :/
    It's a general rule that \ln(e^{x})=x and e^{\ln(x)}=x as they are inverses of each other

    You have \ln(x)=-1

    Raise both sides to the power of e

    This gives:

    Spoiler:
    Show

    e^{\ln(x)}=e^{-1}

    Which means:

    x=e^{-1}=\dfrac{1}{e}
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    (Original post by Robbie242)
    It's a general rule that \ln(e^{x})=x and e^{\ln(x)}=x as they are inverses of each other

    You have \ln(x)=-1

    Raise both sides to the power of e

    This gives:

    Spoiler:
    Show

    e^{\ln(x)}=e^{-1}

    Which means:

    x=e^{-1}=\dfrac{1}{e}
    where has the ln gone? :l
    and the e
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    (Original post by jacksonmeg)
    where has the ln gone? :l
    It's the inverse so both ln and e cancel each other leaving only the f(x) behind in this case f(x)=x
    If you don't understand this you might need to read chapter 3 of your C3 book, it has explanations on this.
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    (Original post by Robbie242)
    It's the inverse so both ln and e cancel each other leaving only the f(x) behind in this case f(x)=x
    If you don't understand this you might need to read chapter 3 of your C3 book, it has explanations on this.
    alright ill take a look at it, can't wait to finish maths, abandon the ship before it sinks
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    (Original post by Robbie242)
    It's the inverse so both ln and e cancel each other leaving only the f(x) behind in this case f(x)=x
    If you don't understand this you might need to read chapter 3 of your C3 book, it has explanations on this.
    the answer is (1/e, -1/e), not sure how to get y, something to do with reflection in y=x? cant find explanation of it in the book
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    (Original post by jacksonmeg)
    is it (1/e, -1/e) ?
    because inverse graphs are reflections of each over in y=x, or is my reasoning completely wrong? xD
    The x-coordinate is correct

    Let me find the y-coordinate

    when x=\dfrac{1}{e}, y=\dfrac{1}{e}\ln(\dfrac{1}{e} )

    Using log laws we have y=\dfrac{1}{e}(\ln(1)-\ln(e))

    \ln(e)=1 so y=-\dfrac{1}{e}

    I'm not sure about reasoning actually but working out the actual coordinate gives the answer the book has
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    (Original post by Robbie242)
    The x-coordinate is correct

    Let me find the y-coordinate

    when x=\dfrac{1}{e}, y=\dfrac{1}{e}\ln(\dfrac{1}{e} )

    Using log laws we have y=\dfrac{1}{e}(\ln(1)-\ln(e))

    \ln(e)=1 so y=-\dfrac{1}{e}

    I'm not sure about reasoning actually but working out the actual coordinate, you are correct!
    not correct! I read the back of the book, I dont understand logs at all
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    (Original post by jacksonmeg)
    not correct! I read the back of the book, I dont understand logs at all
    Then read chapter 3 and do questions from there on logs, there's not point jumping into differentiation if you don't know your logs
 
 
 
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