Find dy/dx by implicit differentiation

[momogiri]

Question:
Find dy/dx by implicit differentiation
4x^2 + 3xy - y^2 = 6

Attempt:
Ok, just a forewarning that I suck at differentiations, limits, what-not, so..
Following my textbook, it says I should Differentiate both sides of the equation
So...
(d/dx)(4x^2 + 3xy - y^2) = (d/dx)(6)
(d/dx)(6) = 0, so I have to solve the derivatives of the other side..
Now, this is the part that I'm unsure about, so please guide me here:
(d/dx)(4x^2) = 8x right? Since I bring down the two and subtract 1
Ok, so.. (d/dx)(3xy), using the chain rule which is f(g(x)) = f'(g(x))*g'(x), I figured:
y(...) is the outer function and 3x is the inner funtion
thus, making it (y(3x))' * (3x)' = (dy/dx)(3x) * 1?

Can someone please clarify that last part I wrote for me? Also, it'd be great if someone can clarify the difference and meaning between dy/dx, d/dx, and d/dy..
Then I'll try and move on XD
Please and thank you for your time XD

PS: I do realize I haven't got through the whole question, but I will, once the above is clarified as correct, etc. :D

 

[bob1182006]

you would use the product rule for 3xy, the chain rule applies to y since y is a function of x.
so you'd have f(x)g(x) = f(x)g'(x)+g(x)f'(x) where either f or g is y. and the derivative of y will be dy/dx.

dy/dx is the derivative of y with respect to x. d/dx and d/dy are operators I think and you use them on a function like y to take the derivative of that function with respect to x or y.

 

[HallsofIvy]

Question:
Find dy/dx by implicit differentiation
4x^2 + 3xy - y^2 = 6

Attempt:
Ok, just a forewarning that I suck at differentiations, limits, what-not, so..
Following my textbook, it says I should Differentiate both sides of the equation
So...
(d/dx)(4x^2 + 3xy - y^2) = (d/dx)(6)
(d/dx)(6) = 0, so I have to solve the derivatives of the other side..
Now, this is the part that I'm unsure about, so please guide me here:
(d/dx)(4x^2) = 8x right? Since I bring down the two and subtract 1

Yes, that is correct.

Ok, so.. (d/dx)(3xy), using the chain rule which is f(g(x)) = f'(g(x))*g'(x), I figured:
y(...) is the outer function and 3x is the inner funtion

No, there is no "inner"or "outer" function- this is not composition of functions, it is just a product of variables- use the product rule: (3xy)'= (3y)(x)'+ (3x)y'= 3y+ 3xy' since you are differentiating with respect to x.

thus, making it (y(3x))' * (3x)' = (dy/dx)(3x) * 1?

Can someone please clarify that last part I wrote for me? Also, it'd be great if someone can clarify the difference and meaning between dy/dx, d/dx, and d/dy..
Then I'll try and move on XD
Please and thank you for your time XD

PS: I do realize I haven't got through the whole question, but I will, once the above is clarified as correct, etc. :D

Okay, I think I did "clarify" above. As for the second question, dy/dx means the derivative of y with respect to x- presumably, y is a function of x- if not that derivative is just 0. d/dx means the derivative of some function with respect to x. Usually that is used when the function is just to long to fit reasonably inside the derivative
[tex]d/dx (x^3- 3sin(x)+ xe^x- 3xcos(x)+ x^3- 3x^2+ 4x)[/itex]
rather than
$$\frac{d (x^3- 3sin(x)+ xe^x- 3xcos(x)+ x^3- 3x^2+ 4x)}{dx}$$

d/dy is the same thing except that you are differentiating with respect to the variable y. Presumably now the function being differentiated is a function of y. If there are any other variables in the function, they should be also functions of y. You probably haven't had "partial derivatives" yet. That's what you would use if you had a function of two or more independent variable.

 

[momogiri]

Oh ok
So for the part 3xy , does the product rule have to be applied twice then?
As in
(3)(x)'+(x)(3)' and then (3x)(y)' + (y)(3x)'?

Edit: No wait, scratch that
It's just (3x)(y)' + (y)(3x)' then, right?

(3xy)'= (3y)(x)'+ (3x)y'= 3y+ 3xy'

I'm not exactly sure how you did that

 

[bob1182006]

yes but you can simplify (3x)'.

for the (3xy)' HallsofIvy just used the product rule, you used it too your answers are exactly the same when you simplify (3x)'.

 

[rocomath]

$$3[xy]$$

$$3[xy'+y]$$

$$3xy'+3y$$

 

[momogiri]

Ok, gotcha :D
I have another quick question, would y' become dy/dx then? That's what dy/dx means, right?

 

[bob1182006]

yes y'=dy/dx

 

[momogiri]

:D Awesome
So in order to find (d/dx)(y^2), now I have to use the chain rule, right?
(...)^2 would be outer and y would be inner, so..

f'(g(x))*g'(x) = ... 2(y)*y' then?
Since the derivative of ^2, I bring down the two, but I don't do anything to the y inside, right?

 

[bob1182006]

right (y^2)' = 2yy'

and yes you can't do anything to y' since y' is a function of x that you don't know since it's defined implicitly.

 

[momogiri]

:D
Ok, so now that I've got the d/dx of the values, I assume I'll have to isolate dy/dx
So..

8x + 3x(dy/dx) + 3y - 2y(dy/dx) = 0
then, putting dy/dx on one side..
2y(dy/dx) - 3x(dy/dx) = 8x + 3y
meaning
(dy/dx)(2y - 3x) = (8x + 3y)
then
(dy/dx) = (8x + 3y)/(2y - 3x)
:D
Do you think I am allowed to leave it like that?

 

[bob1182006]

yep that's right, since you can't reduce the fraction anymore.

 

[momogiri]

YAY :D
Thank you so much bob for sticking with me so long :D :D:D:D:D:D

 

[bob1182006]

no problem, glad I could help

 

[momogiri]

lols, omg.
I'm on my next question, and I'm sort of stuck.. again :(

So here's the next question:
Find dy/dx by implicit differentiation.
x^2y^2 + x sin(y)= 9

Ok, so I know I'm to take d/dx on both sides
But if it's (d/dx) of (x^2y^2), how would I do it?
Would I use the chain rule for x^2 and y^2 separately, and then use the product rule?

 

[bob1182006]

no you would use the product rule first, and when you get to (y^2)' you use the chain rule.

 

[momogiri]

Oh ok, so assuming I did it correctly, it'd be
x^2(y^2)' + y^2(2x)
and (y^2)' = 2y(y)'

I'm unsure about the d/dx of x sin(y).. would I be using the chain rule and then the product rule?

 

[bob1182006]

product then chain rule for (sin y)' since you have xsin(y) you need to be doing the derivative of just sin(y) before you can use the chain rule.

 

[momogiri]

Right!
Ok, so I solved
x(sin y)' + (sin y)(x)'
Which becomes x(cos y*y') + (sin y)
Since (sin y)' = (cos y)(y') and x' = 1
And now, isolating dy/dx...

2yx^2(dy/dx) + x(cos y)(dy/dx) + (sin y) = 0
then 2yx^2(dy/dx) + x(cos y)(dy/dx) = -(sin y)
(dy/dx)(2yx^2 + x(cos y)) = -(sin y)

So (dy/dx) = -(sin y)/(2yx^2 + x(cos y)) right?
I inputted it in my online math homework site place, and it's wrong ToT
What did I do wrong?

 

[bob1182006]

when you took the derivative of x^2y^2 you only have down the 2yy'x^2 part, you're missing the +2xy^2.

 

[momogiri]

Oh man, I can't believe I missed that -__-;;;
But uh, I've started on my next question, and I'm a little unsure (again)
So here's the question

Find dy/dx by implicit differentiation.
√xy = 2 + x2y

So I understand that I have to take d/dx for both sides, so..
I changed squareroot of xy into x^0.5y^0.5
Then I want to use the product rule for x^0.5y^0.5
So it becomes (x^0.5)(y^0.5)' + (y^0.5)(x^0.5)'

The thing that got me is to find (y^0.5)'
I know I have to use the chain rule..
f'(g(x))*(g'(x))
I'm unsure about the f'(g(x)) part..
So I bring down the 0.5, and then subtract 1 at the power??
Would that be 0.5y^-0.5 then? Or.. :(

Edit:
Oh, and for (x^0.5)', would that be 0.5x^-0.5 too?

 

[rocomath]

tbh, i don't think splitting it into 2 radicals is the best way to go.

$$\sqrt{xy}=2+2xy$$

just apply the chain rule & product rule and you'll be fine

 

[momogiri]

Do you mean express d/dx of $$\sqrt{xy}$$ as 0.5(xy)^-0.5?

 

[bob1182006]

yes and then you'd have to multiply that by (xy)'

your result was the same it's just they're separate and keeping them combined will keep the answer a bit simpler.

 

[rocomath]

$$\sqrt{xy}$$

apply the power rule then chain rule of the inner which is a product

$$\frac{1}{2}(xy)^{-{\frac{1}{2}}}(PR)$$ pr = product rule of inner

 

[momogiri]

Ok, so what I did was I got 0.5(xy)^-0.5, then into 0.5(xy'+y)^-0.5
Then I figured out the other side, in which the 2 is gone, and d/dx of x^2y is x^2y' + 2xy

Then.. I'm trying to solve for dy/dx, so..
$$\frac{1}{\sqrt{0.5(xy'+y)}}=x^{2}y' + 2xy$$
Which becomes
$$\frac{1}{\sqrt{0.5(xy'+y)}} - x^{2}y' = 2xy$$
And then... I'm stuck :(

 

[bob1182006]

$$(\sqrt{xy})'\neq\frac{1}{\sqrt{0.5(xy'+y)}}$$

you had it partially right with:
$$\frac{1}{2\sqrt{xy}}$$

you just need to continue using the chain rule since you have (f(g(x)))'=f'(g(x))g'(x) you did f'(g(x)) now you need to multiply by g'(x). where g(x)=xy

 

[momogiri]

So the part I'm missing is the g'(x) of $$g'(x)*\frac{1}{\sqrt{0.5(xy'+y)}}$$?

 

[bob1182006]

no you're missing g'(x) of $$g'(x)*\frac{1}{2\sqrt{xy}}$$

 

[momogiri]

Oh...
Where did the $$\frac{1}{2\sqrt{xy}}$$ come from? XD

And g'(x) = (xy)' ?
I'll have to use the product rule then, right?

 

[bob1182006]

yes, and that fraction is the f'(g(x)) of the function $$\sqrt{xy}$$

since $$(\sqrt{u})'=\frac{u'}{2\sqrt{u}}$$

 

[momogiri]

Oh ok, so $$\frac{1}{2\sqrt{xy}}*(xy'+y)=x^{2}y' + 2xy$$?

 

[bob1182006]

depends, was the right hand side originally $$2+2xy$$ or $$2+x^2y$$?

if it was yx^2 then yes it's right, if it wasn't then no

 

[momogiri]

Yep, it was originally $$2+x^2y$$

So $$\frac{1}{2\sqrt{xy}}*(xy'+y)=x^{2}y' + 2xy$$
becomes
$$\frac{1}{2\sqrt{xy}}xy'+\frac{1}{2\sqrt{xy}}y=x^{2}y' + 2xy$$
which I shuffle..
$$\frac{1}{2\sqrt{xy}}xy' - x^{2}y' = 2xy - \frac{1}{2\sqrt{xy}}y$$
and stuff..
$$y'(\frac{1}{2\sqrt{xy}}x - x^{2}) = 2xy - \frac{1}{2\sqrt{xy}}y$$
thus
$$y' = \frac{2xy - \frac{1}{2\sqrt{xy}}y}{\frac{1}{2\sqrt{xy}}x - x^{2}}$$?

 

[bob1182006]

correct, you could also reduce it a bit more by getting everything with a common denominator of 2sqrtxy to cancel that out.