I was just wondering where the y'/(dy/dx) in implicit differentiation comes from x 2 + y 2 = 25 ( d / d x ) x 2 + ( d / d y ) y 2 ∗ ∗ ( d y / d x ) ∗ ∗ = 25 ( d / d x ) 2 x + 2 y ( d y / d x ) = 0 ( d y / d x ) = − x / yWhere does the bold part come from?

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I was just wondering where the y&#039;/(dy/dx) in implicit differentiation comes from      x    2    +      y    2    =  25  (  d      /    d  x  )      x    2    +  (  d      /    d  y  )      y    2    ∗  ∗  (  d  y      /    d  x  )  ∗  ∗  =  25  (  d      /    d  x  )  2  x  +  2  y  (  d  y      /    d  x  )  =  0  (  d  y      /    d  x  )  =  −  x      /    yWhere does the bold part come from?

Simeon Hester · Accepted Answer

When you implicitly differentiate       x    2    +      y    2    =  25, you are differentiating with respect to a particular variable—in this case, x, so:                              d                      d            x                          (                  x          2                +                  y          2                )                            =                  d                      d            x                          25                                      d                      d            x                          (                  x          2                )        +                  d                      d            x                          (                  y          2                )                            =        0                            2        x        +        2        y                              d            y                                d            x                                              =        0                            2        y                              d            y                                d            x                                              =        −        2        x                                                  d            y                                d            x                                              =        −                  x          y                    From the 3rd line to the 4th line,       d          d      x        (      y    2    ) is the derivative with respect to x of       y    2  , in which (as in Ryan Budney&#039;s comment) we assume that y is some function of x, so we apply the chain rule, differentiating       y    2   with respect to y and multiplying by the derivative of y with respect to x to get   2  y            d      y              d      x      .edit: I think it might be useful if I introduced a slightly different notation: Let       D    x   be the differential operator with respect to x, which you have previously written as       d          d      x       (and, similarly,       D    y   is the differential operator with respect to y). When we apply the differential operator to something, we read and write it like a function:       D    x    (      x    2    )  =  2  x is &quot;the derivative with respect to x of       x    2   is   2  x.&quot;Now, rewriting the work above in this notation:                              D          x                (                  x          2                +                  y          2                )                            =                  D          x                (        25        )                                      D          x                (                  x          2                )        +                  D          x                (                  y          2                )                            =        0                            2        x        +                  D          y                (                  y          2                )                  D          x                (        y        )                            =        0                            2        x        +        2        y                  D          x                (        y        )                            =        0                            2        y                  D          x                (        y        )                            =        −        2        x                                      D          x                (        y        )        =                              d            y                                d            x                                              =        −                  x          y                    And, to your question of finding             d      x              d      y      :                              D          y                (                  x          2                +                  y          2                )                            =                  D          y                (        25        )                                      D          y                (                  x          2                )        +                  D          y                (                  y          2                )                            =        0                                      D          x                (                  x          2                )                  D          y                (        x        )        +        2        y                            =        0                            2        x                  D          y                (        x        )        +        2        y                            =        0                            2        x                  D          y                (        x        )                            =        −        2        y                                      D          y                (        x        )        =                              d            x                                d            y                                              =        −                  y          x

saucletbh · Accepted Answer

In symbols, the chain rule gives:            d      (              y        2            )              d      x        =            d      (              y        2            )              d      y                  d      y              d      x        =  2  y            d      y              d      x

I was just wondering where the y'/(dy/dx) in implicit differentiation comes from. x^2+y^2=25 (d/dx)x^2+(d/dy)y^2∗∗(dy/dx)∗∗=25(d/dx) 2x+2y(dy/dx)=0 (dy/dx)=−x/y Where does the bold part come from? Wikipedia says it's a byproduct of the chain rule, but it's just not clicking for me.

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