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Thread: Implicit Differentiation

  1. #1
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    Implicit Differentiation

    I am learning about implicit differentiation and although I understand the rules, I am having hard time understanding the notation.

    I have failed to understand differentiation with regards the variables i.e. differentiation of a variable with respect to x and y. Lets start with a simple example. I will carry the calculations up to the point where I could ask all my questions:

    (1) 7y4+x3y+x=4
    (2) d/dx[7y4+x3y+x]=d/dx[4]
    (3) d/dx[7y4]+d/dx[x3y]+d/dx[x]=0

    We are at the point where I could ask my first question. Lets isolate the term to the right of the equal sign and assume that it is a separate function y=4 then dy/dx would be 0. What would the derivative be with respect to y? Continuing:

    (4) 28y3[dy/dx]+x3[dy/dx]+3yx2+1=0

    I think here we first assume that y=u and then take du/dx, am I correct? But then why do we continue to carry dy/dx? If we isolate d/dx[7y4] from (3) above, what would be its derivative with regards to y?

    Please, explain as if you were explaining it to a four year old...

    Thanks in advance.
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  2. #2
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    Re: Implicit Differentiation

    Quote Originally Posted by Soundork View Post
    I am learning about implicit differentiation and although I understand the rules, I am having hard time understanding the notation.

    I have failed to understand differentiation with regards the variables i.e. differentiation of a variable with respect to x and y. Lets start with a simple example. I will carry the calculations up to the point where I could ask all my questions:

    (1) 7y4+x3y+x=4
    (2) d/dx[7y4+x3y+x]=d/dx[4]
    (3) d/dx[7y4]+d/dx[x3y]+d/dx[x]=0 So far so good.

    We are at the point where I could ask my first question. Lets isolate the term to the right of the equal sign and assume that it is a separate function y=4 then dy/dx would be 0. What would the derivative be with respect to y? Continuing:

    (4) 28y3[dy/dx]+x3[dy/dx]+3yx2+1=0

    This is correct here's why (taking each term separately):

    d/dx[7y4] = d/dy [7y^4] * [dy/dx] . that's the chain rule
    =28y^3 * [dy/dx]



    d/dx [x^3y] this is a product so you need to use the product rule ...let u= x^3 and v=y
    =uv' + vu' (the derivatives are wrt x)
    =x^3*d/dx(y) + y*d/dx(x^3)
    =x^3 * [dy/dx] + y*3x^2
    = x^3 [dy/dx] + 3x^2y


    The third term is obvious.


    I think here we first assume that y=u and then take du/dx, am I correct? No But then why do we continue to carry dy/dx? If we isolate d/dx[7y4] from (3) above, what would be its derivative with regards to y? In the second term you can assume that u= 7y^4, so you need to find du/dx = du/dy * dy/dx (Chain rule) That's why the dy/dx is still there.

    Please, explain as if you were explaining it to a four year old...

    Thanks in advance.
    see comments in red
    Last edited by Debsta; Sep 25th 2018 at 06:14 PM.
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  3. #3
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    Re: Implicit Differentiation

    Quote Originally Posted by Soundork View Post
    I am learning about implicit differentiation and although I understand the rules, I am having hard time understanding the notation.
    I have failed to understand differentiation with regards the variables i.e. differentiation of a variable with respect to x and y. Lets start with a simple example. I will carry the calculations up to the point where I could ask all my questions:
    (1)$7y^4+x^3y+x=4$
    I have told generations of students: forget fancy notation, just do the derivative.
    (1)$7y^4+x^3y+x=4$
    $28y^3y'+[(3x^2y)+x^3y']+1=0$ Solve for $y'$.
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