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    Where have i gone wrong here ? Name:  1466884406377-1903051015.jpg
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Size:  404.2 KB i tried to check it but my checking suggests whst i have done is wrong...
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    (Original post by coconut64)
    Where have i gone wrong here ? Name:  1466884406377-1903051015.jpg
Views: 108
Size:  404.2 KB i tried to check it but my checking suggests whst i have done is wrong...
    Use the chain rule along with the quotient rule.
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    What you have done wrong is that
    v is equal to (x+3) to the power of minus a half not a half, because it's the denominator, you've written (x+3) to the power of a half, which would just be square root of (x+1) multiplied by square root of (x+3), not 1 over square root of (x+3)
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    u = (x+1)^0.5
    u' = 0.5(x+1)^(-0.5)

    v = (x+3)^(-0.5)
    v' = -0.5(x+3)^(-1.5)

    your v is wrong, remember because the (x+3) is on the bottom it's to a minus power when written not as a fraction
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    Hi! It's that step in the middle where you suddenly get rid of (x+3) from the quotient. It should be:

     \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{ \frac{1}{2}(x+3)^{1/2}(x+1)^{-1/2}-\frac{1}{2}(x+3)^{-1/2}(x+1)^{1/2} }{(x+3)}  = \frac{1}{2}(x+3)^{-1/2}(x+1)^{-1/2}-\frac{1}{2}(x+3)^{-3/2}(x+1)^{1/2}

    You can do the rest!
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    Multiply the numerator and denominator by root[(x+3)(x+1)] and you're good to go, son.

    (Original post by aerofanatic)
    Hi! It's that step in the middle where you suddenly get rid of (x+3) from the quotient. It should be:

     \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{ \frac{1}{2}(x+3)^{1/2}(x+1)^{-1/2}-\frac{1}{2}(x+3)^{-1/2}(x+1)^{1/2} }{(x+3)}  = \frac{1}{2}(x+3)^{-1/2}(x+1)^{-1/2}-\frac{1}{2}(x+3)^{-3/2}(x+1)^{1/2}

    You can do the rest!
    Also that should be -1/2 on the index there, not -3/2 ;D
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    (Original post by RDKGames)
    .
    Also that should be -1/2 on the index there, not -3/2 ;D
    Tested the result on a function plotter; seems okay as it is!
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    (Original post by aerofanatic)
    Tested the result on a function plotter; seems okay as it is!
    Ah damn, apologies, didn't realise you divided through by x+3, took that minus sign as an equals sign to the final answer :/
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    (Original post by aerofanatic)
    Hi! It's that step in the middle where you suddenly get rid of (x+3) from the quotient. It should be:

     \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{ \frac{1}{2}(x+3)^{1/2}(x+1)^{-1/2}-\frac{1}{2}(x+3)^{-1/2}(x+1)^{1/2} }{(x+3)}  = \frac{1}{2}(x+3)^{-1/2}(x+1)^{-1/2}-\frac{1}{2}(x+3)^{-3/2}(x+1)^{1/2}

    You can do the rest!
    Hi, thanks for helping. I have tried this again butstill didnt get it right. I dont get why it is -1/2 not +1/2 since there are two negative signs Name:  1466890876358-1323170548.jpg
Views: 70
Size:  393.7 KB thanks!
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    (Original post by coconut64)
    Hi, thanks for helping. I have tried this again butstill didnt get it right. I dont get why it is -1/2 not +1/2 since there are two negative signs Name:  1466890876358-1323170548.jpg
Views: 70
Size:  393.7 KB thanks!
    You're doing it wrong.

    In the quotient rule, let's say for example in \frac{x+2}{x}, this is of the form \frac{v}{u}.

    You're going to set v = x+2 \Rightarrow \frac{\mathrm{d}v}{\mathrm{d}x} = 1 and u = x \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x}.

    Note that we are not setting u = x^{-1}.

    The whole point of the quotient rule is that you avoid that, so in your answer you should have u = (x+3)^{1/2}.

    This gets you y = \frac{v}{u} = \frac{\sqrt{x+1}}{\sqrt{x+3}} which is good, that's what you want to differentiate.

    If we insist on using your definition, then we get y = \frac{(x+1)^{1/2}}{(x+1)^{-1/2}} = \frac{\sqrt{x+1}}{\frac{1}{\sqrt  {x+3}}} = \sqrt{x+1}\sqrt{x+3}, nothing at all like your original function, you get the drift?
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    (Original post by Zacken)
    You're doing it wrong.

    In the quotient rule, let's say for example in \frac{x+2}{x}, this is of the form \frac{v}{u}.

    You're going to set v = x+2 \Rightarrow \frac{\mathrm{d}v}{\mathrm{d}x} = 1 and u = x \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x}.

    Note that we are not setting u = x^{-1}.

    The whole point of the quotient rule is that you avoid that, so in your answer you should have u = (x+3)^{1/2}.

    This gets you y = \frac{v}{u} = \frac{\sqrt{x+1}}{\sqrt{x+3}} which is good, that's what you want to differentiate.

    If we insist on using your definition, then we get y = \frac{(x+1)^{1/2}}{(x+1)^{-1/2}} = \frac{\sqrt{x+1}}{\frac{1}{\sqrt  {x+3}}} = \sqrt{x+1}\sqrt{x+3}, nothing at all like your original function, you get the drift?
    Thanks, i figured something was wrong with it. I tried using just (x+3) 1/2 but it still doesnt give me the answer. My book says its y=u/v so i dont know if thats why...Name:  14668928939381720090532.jpg
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    (Original post by coconut64)
    Thanks, i figured something was wrong with it. I tried using just (x+3) 1/2 but it still doesnt give me the answer. My book says its y=u/v so i dont know if thats why...Name:  14668928939381720090532.jpg
Views: 66
Size:  387.0 KB
    I agree up to the second = after dy/dx, after that, your simplification is wrong.

    Pay attention to your indices. It should be:

    \displaystyle \frac{1}{2}(x+1)^{-1/2}(x+3)^{1/2 - 1} + \frac{1}{2}(x+3)^{-1/2 - 1}(x+1)^{-1/2}

    Which gets you:

    \displaystyle \frac{1}{2}(x+1)^{-1/2}(x+3)^{-1/2} + \frac{1}{2}(x+3)^{-3/2}(x+1)^{-1/2}

    Now when you factorise, be very careful, pay attention to the indices.
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    (Original post by Zacken)
    I agree up to the second = after dy/dx, after that, your simplification is wrong.

    Pay attention to your indices. It should be:

    \displaystyle \frac{1}{2}(x+1)^{-1/2}(x+3)^{1/2 - 1} + \frac{1}{2}(x+3)^{-1/2 - 1}(x+1)^{-1/2}

    Which gets you:

    \displaystyle \frac{1}{2}(x+1)^{-1/2}(x+3)^{-1/2} + \frac{1}{2}(x+3)^{-3/2}(x+1)^{-1/2}

    Now when you factorise, be very careful, pay attention to the indices.
    Why did you multiply du/dx and dv/dx? I thought it is v* du/dx - u *dv/dx... This is the method i used. Name:  1466893988105-248481546.jpg
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    (Original post by coconut64)
    Why did you multiply du/dx and dv/dx? I thought it is v* du/dx - u *dv/dx...
    Err... I haven't. I'm simply taken what you've written on your second line and did the division correctly for you.

    (x+3)^{-1}\bigg[\frac{1}{2}(x+1)^{-1/2} (x+3)^{1/2} - \frac{1}{2}(x+3)^{-1/2}(x+1)^{1/2}\bigg]
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    (Original post by coconut64)
    Hi, thanks for helping. I have tried this again butstill didnt get it right. I dont get why it is -1/2 not +1/2 since there are two negative signs Name:  1466890876358-1323170548.jpg
Views: 70
Size:  393.7 KB thanks!

    Careful, don't get confused with the notation! It seems as though you could do with an explanation as to how the quotient rule works. lets say  f(x) is what we want to differentiate. It's more conventional to express the quotient as  \frac{u}{v} (which is what I'll do here) rather than  \frac{v}{u} but it really doesn't matter as long as you stay consistent. The quotient rule says that if  f(x)=\frac{u}{v} , then the derivative of  f(x) ,  f'(x) is given by
     f'(x)=\frac{vu'-uv'}{v^2}
    If you're keen to derive this and see why then you can do it from the product rule using  \frac{u}{v}=uv^{-1} .

    If we let  f(x)=\frac{x+1}{x+3} , we see it's already in the form of a quotient (pretty much a fraction), with  u=(x+1)^{1/2} and  v=(x+3)^{1/2} . The derivatives of each of these:
     u'=\frac{1}{2} (x+1)^{-1/2}
     v'=\frac{1}{2} (x+3)^{-1/2}
    Plugging these into the formula for the derivative, we have
      f'(x) = \frac{(x+3)^{1/2}\frac{1}{2}(x+1)^{-1/2}-(x+1)^{1/2}\frac{1}{2}(x+3)^{-1/2} }{((x+3)^{1/2})^2}}
     = \frac{1}{2}(x+3)^{-1/2}(x+1)^{-1/2}-\frac{1}{2}(x+3)^{-3/2}(x+1)^{1/2}
    and after this point, the only thing left to do is to manipulate the answer into the form you need. It all comes with practice!
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    (Original post by aerofanatic)
    Careful, don't get confused with the notation! It seems as though you could do with an explanation as to how the quotient rule works. lets say  f(x) is what we want to differentiate. It's more conventional to express the quotient as  \frac{u}{v} (which is what I'll do here) rather than  \frac{v}{u} but it really doesn't matter as long as you stay consistent. The quotient rule says that if  f(x)=\frac{u}{v} , then the derivative of  f(x) ,  f'(x) is given by
     f'(x)=\frac{vu'-uv'}{v^2}
    If you're keen to derive this and see why then you can do it from the product rule using  \frac{u}{v}=uv^{-1} .

    If we let  f(x)=\frac{x+1}{x+3} , we see it's already in the form of a quotient (pretty much a fraction), with  u=(x+1)^{1/2} and  v=(x+3)^{1/2} . The derivatives of each of these:
     u'=\frac{1}{2} (x+1)^{-1/2}
     v'=\frac{1}{2} (x+3)^{-1/2}
    Plugging these into the formula for the derivative, we have
      f'(x) = \frac{(x+3)^{1/2}\frac{1}{2}(x+1)^{-1/2}-(x+1)^{1/2}\frac{1}{2}(x+3)^{-1/2} }{((x+3)^{1/2})^2}}
     = \frac{1}{2}(x+3)^{-1/2}(x+1)^{-1/2}-\frac{1}{2}(x+3)^{-3/2}(x+1)^{1/2}
    and after this point, the only thing left to do is to manipulate the answer into the form you need. It all comes with practice!
    Thanks i really need more practice and i finally got the right answer at the end
 
 
 
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