Implicit Differentiation (1 Viewer)

[Constip8edSkunk]

need some help with this problem:

Two students A and B were asked 2 fin dy/dx for the curve

x^2/y + y = 3

Student A worked at it directly, using the quotient rule and obtained the answer 2xy/(x^2-y^2)

Student B made life easier by multiplying through by y and then differentiating B obtained the answer 2x/(3-2y)

Has somebody made a mistake or can the two solutions be reconciled.

ok both are correct following their methods, and graphing it gets u a circle where pts where vertical tangents can be produced are (-1.5,1.5) and (1.5,1.5) and hor. tgn. at (0,0) and (0,3) which fits in wif both eqn cept in the original eqn y=/=0 (or atleast i think) as then the x^2/y would b undefined. if this is they case whould that mean (0,0) would not b a pt included in the graph, making the sol B methodically wrong (*y)? ...and how can this b reconciled (what is this question asking u 2 write?)

 

[wogboy]

Student A arrived at the correct answer for dy/dx, it's just that student B forgot to mention that x=/=0 (so student B's answer is incomplete hence not fully correct), since the range of the original function (x^2/y + y = 3) did NOT include y=0, so you cannot define dy/dx where y=0 in this case. If you try subbing in x=0 and y=0 into:

dy/dy = 2xy/(x^2 - y^2), dy/dy will be undefined (0/0) so this proves the point.

but if you try substituting x=0 and y=0 into dy/dx = 2x/(3 - 2y), then you will get dy/dx = 0 which is defined (i.e. not undefined). However, student B should have stated that dy/dx=2x/(3-2y) only if y=/=0. The reason for the exclusion of y=0, is that student B multiplied both sides of the original equations of y, and you cannot multiply both sides of an equation by zero, then get another equation which is equivalent to the original. Hence student B is only allowed to multiply both sides of the original equation by y if y=/=0. If y=0, then new equation will not be equivalent to the old one, so student B's answer is invalidated if y=0.

Just as a side note, the rule above also applies for fractions. If you have a fraction xz/yz, then you can only cancel off z if z is non zero. Otherwise it is undefined. Likewise, if you have a fraction x/y, then it is equal to xz/yz only if z is non zero.

So the two solutions can be reconciled as follows. Just add to student B's answer "where y=/=0". Then you can actually derive student B's anwer from student A's answer to prove it's correct and to properly "reconcile it"

2xy/(x^2 - y^2) (dividing top and bottom by y)
= 2x/(x^2/y - y) (Where y=/=0)
= 2x/(3 - y - y)
= 2x/(3 - 2y) (Where y=/=0)

(since x^/y = 3 - y, from the original equation)

 

[Constip8edSkunk]

thx alot m8