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Can someone check if this C3 dy/dx ques is right? watch

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    find dy/dx of

    y = (x+2)^2 / (x-1)^4

    is the answer:
    (-x^2 - 4x + 4) / (x-1)^5

    Thanks
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    (Original post by kiiten)
    find dy/dx of

    y = (x+2)^2 / (x-1)^4

    is the answer:
    (-x^2 - 4x + 4) / (x-1)^5

    Thanks
    Maple disagrees so unfortunately you got it wrong

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    (Original post by kiiten)
    find dy/dx of

    y = (x+2)^2 / (x-1)^4

    is the answer:
    (-x^2 - 4x + 4) / (x-1)^5

    Thanks
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    (Original post by kiiten)
    find dy/dx of

    y = (x+2)^2 / (x-1)^4

    is the answer:
    (-x^2 - 4x + 4) / (x-1)^5

    Thanks
    http://www.wolframalpha.com/ is your friend

    http://www.wolframalpha.com/input/?i...+%2F+(x-1)%5E4
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    Was i right to use the quotient rule where:

    (u^1v - uv^1 ) / v^2 ?
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    (Original post by B_9710)
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    Name:  Screenshot 2016-11-05 19.50.08.png
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Size:  12.4 KB Maybe im missing something here or just not seeing it at all but how did they get to that last step here?
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    (Original post by kiiten)
    Was i right to use the quotient rule where:

    (u^1v - uv^1 ) / v^2 ?
    Quotient or product, doesn't really matter which as long as your working is consistent. Show your working if you want your error to be pointed out.

    (Original post by kiiten)
    Maybe im missing something here or just not seeing it at all but how did they get to that last step here?
    They put the first term under the same denominator as the second by multiplying top and bottom by (x-1) then just factorised the numerator fully.
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    (Original post by kiiten)
    Name:  Screenshot 2016-11-05 19.50.08.png
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Size:  12.4 KB Maybe im missing something here or just not seeing it at all but how did they get to that last step here?
    They made both fractions have a common denominator, then probably expanded the brackets and simplified.
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    (Original post by RDKGames)
    Quotient or product, doesn't really matter which as long as your working is consistent. Show your working if you want your error to be pointed out.



    They put the first term under the same denominator as the second by multiplying top and bottom by (x-1) then just factorised the numerator fully.
    (Original post by Jasaron)
    They made both fractions have a common denominator, then probably expanded the brackets and simplified.
    If i do that i get this. Ill attach my working (i kinda scribbled some of it out xD)

    NOTE: I made a = x+2 and b= x-1

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    (Original post by kiiten)
    Ifdadasdas

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    Expand the thing you get at the end, (2ab-a^2 . b) and re-factorise, taking the -2 out of the brackets and you'll (probably) get the answer that the computer is giving you.
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    (Original post by kiiten)
    If i do that i get this. Ill attach my working (i kinda scribbled some of it out xD)

    NOTE: I made a = x+2 and b= x-1

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    That's right. Now you can factorise the numerator fully.
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    (Original post by RDKGames)
    That's right. Now you can factorise the numerator fully.
    Yeah I got 2(x+2)(x-1) - 4(x+2)(x+2) which expands to the answer they have.

    Thanks everyone
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    (Original post by RDKGames)
    That's right. Now you can factorise the numerator fully.
    If you derive e^x(x+2)^2

    the calculator says you get (x+4)(x+2)e^x
    Can you divide the whole thing by e^x to leave x^2 + 6x + 8

    If so would the stationary points be -4, 4e^x and -2, e^x
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    (Original post by kiiten)
    If you derive e^x(x+2)^2

    the calculator says you get (x+4)(x+2)e^x
    Can you divide the whole thing by e^x to leave x^2 + 6x + 8

    If so would the stationary points be -4, 4e^x and -2, e^x
    Er... Well if you JUST derive it, then no, you cannot divide an expression - only factor it.

    If you're finding the stationary points and make it equal to 0, then loosely speaking you can indeed just divide by e^x but strictly speaking you need a note as to why you're allowed to take this step.
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    (Original post by RDKGames)
    Er... Well if you JUST derive it, then no, you cannot divide an expression - only factor it.

    If you're finding the stationary points and make it equal to 0, then loosely speaking you can indeed just divide by e^x but strictly speaking you need a note as to why you're allowed to take this step.
    So if the expression is equal to something then you can divide by a factor?

    Does that mean the stationary points could be:
    -4, 4e^x
    -2, e^x
    e^x, ?
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    (Original post by kiiten)
    So if the expression is equal to something then you can divide by a factor?
    Well yeah. These are called equations so you can divide both sides by whatever you want as long as it's not 0.

    It wouldn't make sense to divide x on its own by x because then you get 1 - and this is a different quantity than what x is. If you had x=a then you are allowed to divide by x on both sides if x does not equal 0.

    Does that mean the stationary points could be:
    -4, 4e^x
    -2, e^x
    e^x, ?
    First two yeah, but not the last one because x=e^x is not a solution to the differentiated equation.
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    (Original post by RDKGames)
    Well yeah. These are called equations so you can divide both sides by whatever you want as long as it's not 0.

    It wouldn't make sense to divide x on its own by x because then you get 1 - and this is a different quantity than what x is. If you had x=a then you are allowed to divide by x on both sides if x does not equal 0.



    First two yeah, but not the last one because x=e^x is not a solution to the differentiated equation.
    The differentiated equation is e^x(x+4)(x+2) which i made =0 to find the stationary points. Is e^x not a solution because it doesnt have an x in a bracket like the other 2? (sorry that was badly worded)

    NOTE: I have a bunch of other questions that ive answered but need checking. If i post them here would you be able to check them?
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    (Original post by kiiten)
    The differentiated equation is e^x(x+4)(x+2) which i made =0 to find the stationary points. Is e^x not a solution because it doesnt have an x in a bracket like the other 2? (sorry that was badly worded)
    No. Try solving e^x=0

    NOTE: I have a bunch of other questions that ive answered but need checking. If i post them here would you be able to check them?
    Go for it.
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    (Original post by RDKGames)
    No. Try solving e^x=0



    Go for it.
    lne^x = ln0
    x= ? (invalid).

    Thanks

    1. Find coordinates of y = e^x when the gradient is 3
    is x = ln3 and y =3

    2. Find equation of tangent to y= e^x at (3, e^3)
    y - e^3 = e^3(x - 3)
    y= xe^3 - 2e^3

    3. Equation of normal to y=lnx at (3, ln3)
    y - ln3 = -x(x - 3)
    y = -x^2 + 3x + ln3

    4. find nature of the stationary point x=ln2e dy/dx=2e-e^x
    d2y/dx2 = -e^x
    d2y/dx2 = -e^ln2e
    d2y/dx2 = -2e therefore max. point

    5. Estimate gradient of curve y = (x+lnx)^2 when x=3
    change in y / change in x
    (0.0601 + ln0.0601) / 0.01

    - let me know if anything is unclear or if you want me to take a picture of my actual working. Thanks again
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    (Original post by kiiten)
    lne^x = ln0
    x= ? (invalid).

    Thanks
    Precisely, there is no solution to that. This can easily be seen from remembering the graph of e^x and how it is strictly positive for all values of x (and by strictly positive it means e^x>0)

    1. Find coordinates of y = e^x when the gradient is 3
    is x = ln3 and y =3

    2. Find equation of tangent to y= e^x at (3, e^3)
    y - e^3 = e^3(x - 3)
    y= xe^3 - 2e^3

    3. Equation of normal to y=lnx at (3, ln3)
    y - ln3 = -x(x - 3)
    y = -x^2 + 3x + ln3

    4. find nature of the stationary point x=ln2e dy/dx=2e-e^x
    d2y/dx2 = -e^x
    d2y/dx2 = -e^ln2e
    d2y/dx2 = -2e therefore max. point

    5. Estimate gradient of curve y = (x+lnx)^2 when x=3
    change in y / change in x
    (0.0601 + ln0.0601) / 0.01

    - let me know if anything is unclear or if you want me to take a picture of my actual working. Thanks again
    3. You left your gradient as a variable.

    5. Show your steps to estimation - or give your method of doing so. Where does this info come from?
 
 
 
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