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
d/dx (x^3- 3sin(x)+ xe^x- 3xcos(x)+ x^3- 3x^2+ 4x)[/itex]<br /> rather than<br /> \frac{d (x^3- 3sin(x)+ xe^x- 3xcos(x)+ x^3- 3x^2+ 4x)}{dx}<br /> <br /> 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&#039;t had &quot;partial derivatives&quot; yet. That&#039;s what you would use if you had a function of two or more <b>independent</b> 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&#039;+y]

3xy&#039;+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&#039;+y)}}=x^{2}y&#039; + 2xy
Which becomes
\frac{1}{\sqrt{0.5(xy&#039;+y)}} - x^{2}y&#039; = 2xy
And then... I'm stuck :(

 

[bob1182006]

(\sqrt{xy})&#039;\neq\frac{1}{\sqrt{0.5(xy&#039;+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&#039;(x)*\frac{1}{\sqrt{0.5(xy&#039;+y)}}?

 

[bob1182006]

no you're missing g'(x) of g&#039;(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})&#039;=\frac{u&#039;}{2\sqrt{u}}

 

[momogiri]

Oh ok, so \frac{1}{2\sqrt{xy}}*(xy&#039;+y)=x^{2}y&#039; + 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&#039;+y)=x^{2}y&#039; + 2xy
becomes
\frac{1}{2\sqrt{xy}}xy&#039;+\frac{1}{2\sqrt{xy}}y=x^{2}y&#039; + 2xy
which I shuffle..
\frac{1}{2\sqrt{xy}}xy&#039; - x^{2}y&#039; = 2xy - \frac{1}{2\sqrt{xy}}y
and stuff..
y&#039;(\frac{1}{2\sqrt{xy}}x - x^{2}) = 2xy - \frac{1}{2\sqrt{xy}}y
thus
y&#039; = \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.

 

[momogiri]

You mean by time-sing 2xy and x^2 by 2sqrtxy?

 

[bob1182006]

yes and combining the fractions on the top and the bottom you can cancel out 2sqrt(xy) and have just 1 denominator/numerator.

 

[momogiri]

Hmm.. I'm not sure if I did it correctly though, so here it is:
\frac{2xy - \frac{1}{2\sqrt{xy}}y}{\frac{1}{2\sqrt{xy}}x - x^{2}}
into
\frac{\frac{2xy(2\sqrt{xy})}{(2\sqrt{xy})} - \frac{y}{2\sqrt{xy}}}{\frac{x}{2\sqrt{xy}} - \frac{x^{2}(2\sqrt{xy})}{(2\sqrt{xy})}}
and into
\frac{\frac{2xy(2\sqrt{xy}) - y}{2\sqrt{xy}}}{\frac{x - x^{2}(2\sqrt{xy})}{2\sqrt{xy}}}
which becomes
\frac{2xy(2\sqrt{xy}) - y}{x - x^{2}(2\sqrt{xy})}
and..
\frac{4xy(\sqrt{xy}) - y}{x - x^{2}(2\sqrt{xy})}?
I donno if I can simplify it more XP

 

[bob1182006]

yes that's about as simplified as you can go

 

[momogiri]

Man, that was a pain.. but.. I got stuck again :P Surprise surprise..

Question:
Find dy/dx by implicit differentiation.
e^{x/y} = 2x - y

Attempted:
Isn't (d/dx)e^{x/y} = e^{x/y}?
Well, I went ahead anyway, and figured out
(d/dx)2x = 2 and (d/dx)y = y&#039;
Then I um.. rearranged it so it's
y&#039; = 2 - e^{x/y}
Which is wrong. lol

I'm assuming that (d/dx)e^{x/y} \neq e^{x/y} :P Then how do I find it?

 

[bob1182006]

(e^u)&#039;=u&#039;e^u

so you need to use the quotient rule on x/y and multiply that to e^(x/y)

 

[momogiri]

Damn, lol
I'll try it XD

Dang, I thought I had it right, but I got it wrong :P
Here's what I did..
Quotient rule = \frac{g(x)f&#039;(x)-f(x)g&#039;(x)}{(g(x))^{2}}
So I plugged in the numbers as usual..
\frac{(y)(x&#039;)-(x)(y&#039;)}{(y)^{2}} which got me \frac{(y)-(xy&#039;)}{y^{2}}
Therefore the whole thing becomes
\frac{y-xy&#039;}{y^{2}}e^{\frac{x}{y}} = 2 - y&#039;
So I got rid of y^2
y-e^{\frac{x}{y}}xy&#039; = 2y^{2} - y^{2}y&#039;
then shuffled
y-2y^{2} = e^{\frac{x}{y}}xy&#039; - y^{2}y&#039;
Thus
y-2y^{2} = y&#039;(e^{\frac{x}{y}}x - y^{2})
Therefore
y&#039; = \frac{y-2y^{2}}{e^{\frac{x}{y}}x - y^{2}}...?

What did I do wrong this time :/

 

[bob1182006]

when you got rid of y^2 you didn't keep the parenthesis:
(y-xy')e^(x/y) =/=y-e^(x/y)xy'

so you're just missing a e^(x/y) multiplied to y in the numerator

 

[momogiri]

Eeek, I don't see it..
And by getting rid of y^2, I actually meant timesing it to the other side XD

 

[bob1182006]

\frac{(y-xy&#039;)e^{\frac{x}{y}}}{y^2}\neq y-e^{\frac{x}{y}}xy&#039;
and the y^2 being carried over to the other side.
it should be:

ye^{\frac{x}{y}}-xy&#039;e^{\frac{x}{y}}

and continuing from there you are only missing that e^(x/y) on your final answer

 

[momogiri]

Thanks for that, bob :D
I've got another question :/
Use implicit differentiation to find an equation of the tangent line to the curve at the point (0, -0.5).
x^{2} + y^{2} = (2x^{2} + 2y^{2} - x)^{2}

Ok, so x^{2} + y^{2} = (2x^{2} + 2y^{2} - x)^{2} becomes x^{2} + y^{2} = 2x^{4} + 2y^{4} - x^{2}
Which I solved for the d/dx
2x + 2yy&#039; = 8x^{3} + 4(2y)^{3}*2y&#039; - 2x
Shuffling..
2yy&#039; - 4(2y)^{3}*2y&#039; = 8x^{3} - 4x
and again
y&#039;(2y - 8(2y)^{3}) = 8x^{3} - 4x
thus
y&#039; = \frac{8x^{3} - 4x}{2y - 8(2y)^{3}} which equals
y&#039; = \frac{8x^{3} - 4x}{2y - 64(y)^{3}}

Ok. So that's my slope for the tangent line..
And my equation for tangent line is
y = \frac{8x^{3} - 4x}{2y - 64(y)^{3}}x
and at point (0,-0.5)
y + 0.5 = \frac{8x^{3} - 4x}{2y - 64(y)^{3}}x
Making it
y = \frac{8x^{3} - 4x}{2y - 64(y)^{3}}x - 0.5
So I've managed to solve up to here, and.. frankly, I don't know if I'm right XD
Oh, and this is an image they gave me

 

[bob1182006]

no you have (2x^2+2y^2-x)^2=(2x^2+2y^2-x)(2x^2+2y^2-x)
so you have to multiply that out and then you can do the derivative.

or you can do it straight away (u^n)&#039;=nu^{n-1}u&#039;
where u=2x^2+2y^2-x

also in order to find the actual slope of the function at (0,-0.5) when you isolate y' plugin (0,-0.5) that way you'll get a number.
so your equation for the line shoudl be: y+0.5=mx

 

[momogiri]

Oh XD Ok

(Off-topic) Hey Bob, will you be on later tonight?
I have to go eat dinner and travel back to my dorm :/ That'll take an hour or two, so...
It'd be great if you're still available to help me after XD

Edit: Daaang, just read your message XD
Oh well, I'll manage somehow :D I appreciate all your help :D

 

[bob1182006]

sorry but I'm sure other people will be able to help too,
just remember that when you need to do product/quotient rule and then do the chain rule if you have (y^n)' where n is different from 1.