Could someone explain chain rule when it comes to implicit differentiation?so basically I want to know why when we have something like: v ( x ) = x − y + 1If we take the derivative with respect to x, it yields: v ′ ( x ) = 1 − d y d x Now I still don't understand why when it comes to implicit differentiation, we need to tag a y ′ or d y d x after every time we take the derivative of a y term.

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Could someone explain chain rule when it comes to implicit differentiation?so basically I want to know why when we have something like:  v  (  x  )  =  x  −  y  +  1If we take the derivative with respect to x, it yields:      v    ′    (  x  )  =  1  −            d      y              d      x      Now I still don&#039;t understand why when it comes to implicit differentiation, we need to tag a       y    ′   or             d      y              d      x       after every time we take the derivative of a y term.

Tirioliwo · Accepted Answer

Basically when we are &quot;taking differentiation with respect to   x&quot;, we mean that (intuitively) &quot;when   x has a small change, how will the function change&quot;.Now   y can be a function of   x, e.g.   y  =      x    2   or   y  =      e                  e        x            . So a small change in   x will cause a change (called             d      y              d      x      ) in   y.For example, if you are dealing with ddxy2, you need to do it as &quot;if I change slightly   y, we will get 2  y as the change (      d          d      y            y    2    =  2  y). But since we are doing it with respect to   x so we need to multiply the term by &#039;when we change   x a bit, how   y will be changed [            d      y              d      x        =      y    ′  ′]. Hence the answer is   2  y            d      y              d      x      .&quot;Of course it is only intuitively speaking, and the formal proof of chain rules etc. can be found online easily. I am trying to clarify why the &quot;correction&quot; term             d      y              d      x       is necessary.

Deon Moran · Accepted Answer

y is a function of   x so when   x varies, it is going to cause   y to vary along with it.If you want to treat   y as a variable that is independent from   x that is called partial differentiation, and it would give the same result as setting             d      y              d      x        =  0

so basically I want to know why when we have something like: v(x)=x−y+1

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