I have seen how people implicitly differentiate the equation x 2 + y 2 = c. d / d x ( x 2 ) + d / d x ( y 2 ) = d / d x ( c )treating " y" as " f ( x )" and using the chainrule we get 2 x + 2 y ( y ′ ) = 0and solving for y ′ y ′ = − 2 x / 2 yThe problem is that I just don´t understand implicit differentiation, I do know the rules but they don´t make any sense to me. The fact that it is valid to differentiate both " x" and " y" on the same side of the equation is what´s bothering me and even if I see " y" as a function of " x" I just end up imagining x 2 + ( − x 2 + c ) = cwhich doesn´t help me. I also don´t know very much about partial derivatives but I´m willing to learn about them if that helps me understand implicit differentiation.

Question

I have seen how people implicitly differentiate the equation       x    2    +      y    2    =  c.  d      /    d  x  (      x    2    )  +  d      /    d  x  (      y    2    )  =  d      /    d  x  (  c  )treating &quot;  y&quot; as &quot;  f  (  x  )&quot; and using the chainrule we get  2  x  +  2  y  (      y    ′    )  =  0and solving for       y    ′        y    ′    =  −  2  x      /    2  yThe problem is that I just don´t understand implicit differentiation, I do know the rules but they don´t make any sense to me. The fact that it is valid to differentiate both &quot;  x&quot; and &quot;  y&quot; on the same side of the equation is what´s bothering me and even if I see &quot;  y&quot; as a function of &quot;  x&quot; I just end up imagining      x    2    +  (  −      x    2    +  c  )  =  cwhich doesn´t help me. I also don´t know very much about partial derivatives but I´m willing to learn about them if that helps me understand implicit differentiation.

Tiberlaue · Accepted Answer

Maybe this helps. Take the function   f  (  x  )  =      x    2   and some other functions   g  (  x  ) and   h  (  x  ).Let&#039;s differentiate the expression  f  (  x  )  +  (  g  (  x  )      )    2    =  h  (  x  )with respect to   x (not: &quot;differentiate   x&quot;).We use the fact that you can differentiate each summand individually, and the chain rule for   (  g  (  x  )      )    2  , to get      f    ′    (  x  )  +  2  g  (  x  )      g    ′    (  x  )  =      h    ′    (  x  )  .In your case,   f  (  x  )  =      x    2  ,   g  (  x  )  =  y  (  x  ), and   h  (  x  )  =  c, i.e.:  2  x  +  2  y  (  x  )      y    ′    (  x  )  =  0  .Sometimes, the argument   (  x  ) is omitted -- that&#039;s what you have there; but that&#039;s just notation (no deeper meaning, afaik).

Aarav Atkins · Accepted Answer

We are just using the chain ruleIf I said:  f  (  x  )  =  (  2  x  +  1  )    ⟹              d      f              d      x        =  2    g  (  x  )  =      x    2      ⟹              d      g              d      x        =  2  x        d          d      x        (  g  ∘  f  )  (  x  )  =            d      g              d      f                  d      f              d      x        =  2  (  2  x  +  1  )  (  2  )Do have a problem with that?So,   y is a function of   x.       y    2   is a function of   y      d          d      x            y    2    =      d          d      y        (      y    2    )            d      y              d      x        =  2  y            d      y              d      x      And now for the last little bit:      d          d      x        (      y    2    =  c  −      x    2    )    ⟹    2  y            d      y              d      x        =  −  2  xIs that so different from      d          d      x        (      x    2    +      y    2    =  c  )        d          d      x        (      x    2    )  +      d          d      x        (      y    2    )  =      d          d      x        c    2  x  +  2  y            d      y              d      x        =  0              d      y              d      x        =  −      x    y

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