Thread: 2 simple problems concerning implicit differentiation

1. 2 simple problems concerning implicit differentiation

1) Find dy/dx for $4y^2 - xy = 2$

The given answer is; $y / (8y - x)$ but I don't understand how to get that. My procedure was;

$4y^2 - xy = 2$
$8y - (x + y) = 0$
$8y - x - y = 0$
And according to how the answer was gotten, that means I should move over the -y to the other side, followed by the 8y - x together. What I don't understand is why you can't subtract the 8y and -y first to make -7y and then move it over? And why is the 8y - x supposed to be moved together to the other side instead of separately?

2) Find dy/dx for $(x + y) / (2x - y) = 1$

I'm assuming you have to apply the quotient rule first, but can anyone tell me how it's supposed to look after the rule's been applied? I'm doing something wrong in my calculations.

2. Originally Posted by Archduke01
1) Find dy/dx for $4y^2 - xy = 2$

The given answer is; $y / (8y - x)$ but I don't understand how to get that. My procedure was;

$4y^2 - xy = 2$
$8y - (x + y) = 0$
$8y - x - y = 0$
And according to how the answer was gotten, that means I should move over the -y to the other side, followed by the 8y - x together. What I don't understand is why you can't subtract the 8y and -y first to make -7y and then move it over? And why is the 8y - x supposed to be moved together to the other side instead of separately?

2) Find dy/dx for $(x + y) / (2x - y) = 1$

I'm assuming you have to apply the quotient rule first, but can anyone tell me how it's supposed to look after the rule's been applied? I'm doing something wrong in my calculations.
You have to use the chain rule to differentiate. The derivative of $4y^2$ is $8yy'$ because you're differentiating with respect to $x$. You wrote $8y$, but this is just the derivative of $4y^2$ with respect to $y$. So your equation should have been:

$8yy'-(y+xy')=0$

$8yy'-y-xy'=0$

$y'(8y-x)=y$

$y'=\frac{y}{8y-x}$

3. Originally Posted by Archduke01

2) Find dy/dx for $(x + y) / (2x - y) = 1$

I'm assuming you have to apply the quotient rule first, but can anyone tell me how it's supposed to look after the rule's been applied? I'm doing something wrong in my calculations.
Yes, you just have to apply the quotient rule. This is how the rule is applied:

$\frac{(2x-y)\frac{d}{dx}(x+y)-(x+y)\frac{d}{dx}(2x-y)}{(2x-y)^2}=0$

Now you just find the two derivatives in the numerator, and substitute them.

Yes, you just have to apply the quotient rule. This is how the rule is applied:

$\frac{(2x-y)\frac{d}{dx}(x+y)-(x+y)\frac{d}{dx}(2x-y)}{(2x-y)^2}=0$

Now you just find the two derivatives in the numerator, and substitute them.
Am I on the right path if I get $(2x - y)(1 + y') - (x+y)(2-y') / (2x-y)^2 = 0$?

5. Originally Posted by Archduke01
Am I on the right path if I get $(2x - y)(1 + y') - (x+y)(2-y') / (2x-y)^2 = 0$?
Yup

6. $(2x - y)(1 + y') - (x+y)(2-y') / (2x-y)^2 = 0$

I don't know how to proceed from there. I expanded the terms and ended up with ...

$2x + 2xy' - y - yy' - (2x - xy' + 2y - yy') / (2x - y)^2 = 0$
$2xy' + 3xy' - 3y / (2x-y)^2 = 0$

Seems like I messed up somewhere along the line. Could someone please point it out?

7. Originally Posted by Archduke01
$(2x - y)(1 + y') - (x+y)(2-y') / (2x-y)^2 = 0$

I don't know how to proceed from there. I expanded the terms and ended up with ...

$2x + 2xy' - y - yy' - (2x - xy' + 2y - yy') / (2x - y)^2 = 0$
$2xy' + 3xy' - 3y / (2x-y)^2 = 0$

Seems like I messed up somewhere along the line. Could someone please point it out?
Just set the numerator equal to zero, and then expand. You don't need to worry about the denominator. Remember that the denominator can't be
zero. Just solve this for $y'$:

$(2x - y)(1 + y') - (x+y)(2-y')=0$

Just set the numerator equal to zero, and then expand. You don't need to worry about the denominator. Remember that the denominator can't be
zero. Just solve this for $y'$:

$(2x - y)(1 + y') - (x+y)(2-y')=0$
Thanks but I know, that's what I did. I ended up with $2xy' + 3xy' - 3y = 0$

9. Originally Posted by Archduke01
Thanks but I know, that's what I did. I ended up with $2xy' + 3xy' - 3y = 0$

$(2x - y)(1 + y') - (x+y)(2-y')=0$

$2x-y+2xy'-yy'-[2x-xy'+2y-yy']=0$

$-y+2xy'-2y+xy'=0$

$3xy'-3y=0$

$y'=\frac{y}{x}$

You made a simple algebraic error, but you were close.

$(2x - y)(1 + y') - (x+y)(2-y')=0$

$2x-y+2xy'-yy'-[2x-xy'+2y-yy']=0$

$-y+2xy'-2y+xy'=0$

$3xy'-3y=0$

$y'=\frac{y}{x}$

You made a simple algebraic error, but you were close.
Thanks, I corrected it and also got
$3xy'-3y=0$ but the thing is, the given answer is 1/2.

11. Originally Posted by Archduke01
Thanks, I corrected it and also got
$3xy'-3y=0$ but the thing is, the given answer is 1/2.
I reviewed the problem, I can't see anything wrong with it. Did you state the entire problem in your original post? Perhaps you were asked to find $\frac{dy}{dx}$ at the point (2,1)

I reviewed the problem, I can't see anything wrong with it. Did you state the entire problem in your original post? Perhaps you were asked to find $\frac{dy}{dx}$ at the point (2,1)
Nope. It only asked to find dy/dx. Maybe we were supposed to do something with the denominator?

13. Wait a minute, lol, hold on!

$\frac{x+y}{2x-y}=1$

$x+y=2x-y$

Forget the quotient rule!!!

$1+y'=2-y'$

$y'=\frac{1}{2}$

Wait a minute, lol, hold on!

$\frac{x+y}{2x-y}=1$

$x+y=2x-y$

Forget the quotient rule!!!

$1+y'=2-y'$

$y'=\frac{1}{2}$

Oh damn LOL it was as simple as that!

Anyway, thank you so much for your time and sorry for all the questions.