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#1 2018-12-07 09:48:35

Registered: 2018-12-06
Posts: 3

Problem in implicit differentiation

In the first example of implicit differentiation topic ( you are differentiating x² + y² = r² with respect to x. According to partial derivative article
(, to find derivative with respect to x we should treat all other variables as constant then why only r² is treated as constant in the example? Why not y²? Also why we cannot find derivative of y² using power rule? Why you use chain rule only for y²? Last few topics of calculus are hard to understand. Please help me...


#2 2018-12-07 21:24:06

bob bundy
Registered: 2010-06-20
Posts: 8,524

Re: Problem in implicit differentiation

hi Malubia

On the page about partial differentiation f is a function of y and x.  So if you wanted to make a graph of this function a sheet of 2 dimensional graph paper wouldn't be enough.  You need a 3 dimensional graph with f having its values plotted in the 'z' direction.  Imagine you have the x and y axes on a flat surface and the z axis goes up in the air.  If you choose values of x and y, there will be values of z making a (wavy) surface above (and below) the x-y plane.

For partial differentiation we assume that one of the variables (let's say y) is fixed.  That means we are only considering a part of the surface where x and z can vary but y stay constant.  This is now a 2 dimensional problem once more so we can use the usual rules of differentiation. Let's imagine a possible situation where we could use this.  If the surface has a dip down to a lowest point, taking y as constant will be like taking a slice through the surface and making a curve which has a lowest point.  If we now re-do the differentiation with x constant there will similarly be a lowest point.  For each there will still be a variable (the one that we've been pretending is constant) in the equations.  But with a pair of such equations we can find the single point where both x and y are lowest together, in other words the bottom of the dip. 

In the same way we can find maximum points on the surface and also saddle points where one variable is lowest and the other highest.  If you are walking towards a mountain pass, that is an example of a saddle point.  As you go through the pass the ground slopes down in front and behind so you are at a maximum in that direction; but to either side the mountain sides slope up, so in that direction you are at a minimum.  In both directions the partial derivatives are zero.

Now back to the 2 dimensional cases.  If the function is expressed as y = function of x, then the differentiation is relatively straight forward.  But sometimes we don't have that; we may have an equation such as x^2 + y^2 = R where neither variable is the subject of the equation.  In this case it is possible to re-arrange the equation so that y is the subject, but it's not an easy differentiation then, and sometimes you cannot do the re-arrangement at all.

That's when implicit differentiation becomes useful.  Whenever you are differentiating it is always with respect to a variable, let's say x.  Do the x containing components as usual but you have to remember that y is a variable too.  So it must be differentiated with respect to x as well.  And that's the process of implicit differentiation.

It's worth noting that even with a simple function like y = x^2 you are actually using implicit differentiation.  The right hand side becomes 2x.  The left hand side becomes dy/dx.  So it has also been differentiated with respect to x.

Hope that helps,


Thanks for the good wishes in the other post.  Let me know if you discover how to achieve it smile Living for a 1000 years that is.

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob Bundy smile


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