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Math Help - Differentiation y w.r.t.x

  1. #1
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    Differentiation y w.r.t.x

    x^-1 + y^-1 = e^y

    My textbook has a pretty rubbish explanation of how to differentiate y, from what i gather its like x but you stick a dy/dx after it and use the product rule for xy terms, can anyone do the above question with a slightly better explanation than ive given?!
    Thanks!
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  2. #2
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    Quote Originally Posted by darksupernova View Post
    x^-1 + y^-1 = e^y

    My textbook has a pretty rubbish explanation of how to differentiate y, from what i gather its like x but you stick a dy/dx after it and use the product rule for xy terms, can anyone do the above question with a slightly better explanation than ive given?!
    Thanks!
    Hi darksupernova,

    i guess you mean when you differentiate both sides with respect to x,
    and y is a function of x.

    Then you use \frac{d}{dx}f(y)=\frac{dy}{dy}\frac{d}{dx}f(y)=\fr  ac{dy}{dx}\frac{d}{dy}f(y)

    You can just think of swopping denominators, as in

    \left(\frac{10}{3}\right)\frac{6}{5}=\frac{60}{15}  =\left(\frac{10}{5}\right)\frac{6}{3}

    It's called the chain rule

    \frac{d}{dx}x^{-1}+\frac{dy}{dx}\frac{d}{dy}y^{-1}=\frac{dy}{dx}\frac{d}{dy}e^y

    and continue
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  3. #3
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    Thankyou, its a bit odd to get my head round but thats probably because ive never done it!

    Thanks for your help
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  4. #4
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    Hi darksupernova,

    i forgot to mention,

    you can differentiate y with respect to y,
    we can differentiate x with respect to x.

    when we have \frac{d}{dx}f(y)

    and we don't have y expressed in terms of x,
    we've got a dx where we'd prefer to have a dy

    We can multiply any value by 1 without changing it,
    and there are an infinite number of ways to write 1.

    We get dy where we want it by multiplying by \frac{dy}{dy}
    and swopping denominators.
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  5. #5
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    okay i get that part, but this question has thrown me a little...

    Find \frac{dy}{dx} as a function of x if y^2 = 2x + 1.

    How do i make dy/dx a duction of x?
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  6. #6
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    Quote Originally Posted by darksupernova View Post
    okay i get that part, but this question has thrown me a little...

    Find \frac{dy}{dx} as a function of x if y^2 = 2x + 1.

    How do i make dy/dx a duction of x?
    y^2 = 2x + 1
    Differentiating this you get
    2y*dy/dx = 2. Or
    dy/dx = 1/y = 1/sqrt(2x + 1)
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  7. #7
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    Quote Originally Posted by darksupernova View Post
    okay i get that part, but this question has thrown me a little...

    Find \frac{dy}{dx} as a function of x if y^2 = 2x + 1.

    How do i make dy/dx a function of x?
    y^2=2x+1

    \frac{d}{dx}y^2=\frac{d}{dx}(2x+1)=2

    \frac{dy}{dx}\frac{d}{dy}y^2=2

    now y^2 can be differentiated wrt y

    \frac{dy}{dx}2y=2

    \frac{dy}{dx}=\frac{2}{2y}=\frac{1}{y}

    You get it by differentiating the x terms wrt x and the y terms wrt y,

    then dividing by the multiplier of \frac{dy}{dx}
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  8. #8
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    i differentiated it correctly but i didnt substitue y^2 = 2x + 1 back in :S is that because it says as a function of x?
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  9. #9
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    Quote Originally Posted by darksupernova View Post
    i differentiated it correctly but i didnt substitue y^2 = 2x + 1 back in :S is that because it says as a function of x?
    yes, as a function of x, then continue as sa-ri-ga-ma showed.

    y^2=2x+1\ \Rightarrow\ y=\pm\sqrt{2x+1}

    Initially, you could also write

    y=\pm\sqrt{2x+1}

    and calculate

    \frac{dy}{dx}=\pm\frac{d}{dx}(2x+1)^{0.5}=\pm(2)(0  .5)(2x+1)^{-0.5}
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  10. #10
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    It should be
    (0.5)(2)(2x + 1)^-0.5
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  11. #11
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    yeah i got it now, thanks guys, it was just the fact i hadnt substituted in the value for y. Got it now, thank you both!
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