Can someone please help clarify Implicit Differentiaton?

Question

Okay like i get you first find the derivative of the terms provided, but where do you put dy/dx? is it when you find the derivative of y you automatically put dy/dx to be mulitplied to it right? like for example.......xy+y^2=2 get x(dy/dx)+y+2y(dy/dx)=0 after isolating it and everything i get -y/(2y+x) is this right?
IF soo.... Im asked to find the second derivative of it...so after i use the quotient rule i get (2y+x)(-1)-(y+1)(-y)(dy/dx)/(2y+x)^2 if thats right how come i put dy/dx by the derivative of g(x)? can someone please help me out? i really really appreciate it

Manuka when you did this = [(y / (2y+x)) (2y+x) + y (-2y / (2y+x) + 1)] / (2y+x)^2
you forgot to mulitply y and x when you distributed y(2y+x) which would give you xy -2y^2/(2y+x)^2 as a final answer correct??
= [2y - 2y^2 / (2y+x)] / (2y+x)^2

Afk you found the first derivative incorrectly... when you put 2y dy/dx on the right with y you forgot that it is being MULTIPLIED so therefore you cannot just subtract it to the other side. you have to divide it to put it on the side with y.

Answer 1

Yes, you're right. The reason we do this is because of the chain rule: if we have a function f(y), then df/dx = df/dy . dy/dx. So you differentiate it with respect to y and then multiply by dy/dx.
So we have dy/dx = -y / (2y+x), and we want d^2y/dx^2.

Implicit differentiation says
d^2y/dx^2 = [-dy/dx (2y+x) + y (2dy/dx + 1)] / (2y+x)^2
which is where you seem to have slipped up a little. Remember you're differentiating with respect to x, not y. I also don't see where the (y+1) comes from. Note that when you're differentiating something like 2y + x, the dy/dx only attaches to the first part, not the second; so it's 2dy/dx + 1, not (2 + 1)dy/dx or anything like that.

Now we have to substitute in the function we know for dy/dx, so it'll get messy:
d^2y/dx^2 = [-dy/dx (2y+x) + y (2dy/dx + 1)] / (2y+x)^2
= [(y / (2y+x)) (2y+x) + y (-2y / (2y+x) + 1)] / (2y+x)^2
= [2y - 2y^2 / (2y+x)] / (2y+x)^2
= [2y (2y+x) - 2y^2] / (2y+x)^3
= 2y (y+x) / (2y+x)^3.

Answer 2

First of all Implicit mean Hidden and in this context it's Hidden Differentiation of "Y" and "Y" only unless they ask you otherwise. So in other words your only suppose to find the hidden value of "Y" not "X", OK . kool

Secondly you have to remember the importance of product rule, when ever two unknown variables are multiplied you have to use and I am sure you remember the following
F = u ' v + v u ' (product rule)
so first apply that on xy in question
f (X) = xy + y^2 = 2
f' (x) = ( (1 *y) + (x * 1 dy/dx) ) + 2y dy/ dx = 0

( y + x dy/dx ) + 2y dy/dx = 0 and now - y from both sides
x dy/dx + 2y dy/dx = -y and now take common factor of dy/dx

dy/dx ( x +2y) = -3y and divide both sides from (x+2y)

dy/dx = (-3y)/ (x+2y) or in other words Y ' (x) = -3y / (x +2y)

And after this just go ahead and apply quotient rule to find
Y'' (X) .
I hope this Helps and always remember to use the product rule when multiplying two different variables. ok , kool.

P.S chose my answer as the best :) and I will help you in future as well :) and "hhh" thank you for that nudge but still that is how you do implicit functions and comprehension of concept is extremely important, which what I tried to do here :) besides its you not me that has to pass that calculus 1, and I am sure you know the answer for that ....

And also this messages is for "Scarlet Manuka" your incorrect because the first prime is not correct and therefore your answer is not correct either. Although that is the way it's done if you had the right answer :)

Answer 3

Good job on the first... You just need to remember your differentiation rules. If you need the derivative of y, you use dy/dx instead of 1, like when it's x.
I see on the second derivative you used -1 for the derivative of negative y. Just don't forget to use dy/dx! You're doing fine.