Determine y' for xy = 1 by implicit differentiation, Mathematics

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Determine y′ for xy = 1 .

Solution : There are in fact two solution methods for this problem.

Solution 1: It is the simple way of doing the problem.  Just solve for y to obtain the function in the form which we're utilized to dealing with and then differentiate.

y = 1/x ⇒             y′ = - 1/x2

Hence, that's easy sufficient to do.  However, there are some functions for which it can't be done. That's where second solution method comes to play.

Solution 2 (through implicit differentiation):

In this we're going to leave the function in the form which we were given & work with it in that form.  Though, let's recall from the first part of this solution that if we could solve out for y then we will get y like a function of x.  In other terms, if we could solve out for y (as we could in this case, however won't always be capable to do) we get y = y (x).  Let's rewrite the equation to note down this.

                                          xy = x y ( x ) = 1

Be careful here and note down that while we write y ( x ) we don't mean y times x.  What we are noting at this time is that y is some (probably unknown) function of x. It is important to recall while doing this solution technique.

In this solution the next step is to differentiate both sides w.r.t. x as follows,

                                 d ( x y ( x ))/ dx = d (1)/ dx

The right side is simple.  It's just the derivative of constant. The left side is also simple, but we've got to identify that we've in fact got a product here, the x and they ( x ) .  Thus to do the derivative of the left side we'll have to do the product rule.  By doing this gives,

 (1) y ( x ) + x d ( y ( x )) /dx= 0

Now, recall that we have the given notational way of writing the derivative.

d ( y ( x )) / dx = dy/ dx = y′

By using this we get the following,

y + xy′ = 0

Note as well that we dropped the ( x ) on the y as it was just there to remind us that the y was a function of x & now that we've taken the derivative it's no longer needed really. We just desired it in the equation to identify the product rule while we took the derivative.

thus, let's now recall just what were we after. We were after the derivative,  y′ , and notice that there is now a  y′ in the equation.  Thus, to get the derivative all that we have to do is solve the equation for  y′ .

                                                                   y′ = - y/ x

There it is. By using the second solution technique it is our answer. It is not similar with the first solution however. Or at least it doesn't look like the similar derivative that we got from the first solution.  However, recall that we actually do know what y is in terms of x and if we plug that in we will get,

                                            y′ = -       (1/x) /x= -1/ x2

that is what we got from the first solution.  Regardless of the solution technique utilized we should get the same derivative.


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