Question

Differentiation Problem (4)...

Find \(\cfrac{dy}{dx}\) for the equation\(x^2y^3 + y\sin x = 0\)

Answer 1

I know that we have to use implicit differentiation here.

Answer 2

And, yes, I just studied it.

Answer 3

Good start !

Answer 4

So, before anyone wastes his/her time, I will write that out.

Answer 5

On differentiating that, I get :

\(2x y^3 + x^2 \left( 3y^2 \cfrac{dy}{dx} \right) + y \cos x + \cfrac{dy}{dx} \sin x = 0 \)

Answer 6

Or just using the shortcut :

\(\cfrac{dy}{dx} = \cfrac{-(2xy^3 + y\cos x) }{3x^2 y^2 + \sin x}\)

The shortcut I used :

\(\cfrac{dy}{dx} = \cfrac{\text{Derivative of function w.r.t x (taking y as constant)}}{\text{Derivative of function w.r.t y( taking x as constant) }}\)

Answer 7

Am I correct till now?

Answer 8

Yep, you're correct!

Answer 9

Okay, I think I am getting something now :

From the question :

\(x^2 y^3 + y\sin x = 0\)

\(x^2 y^3 = -y \sin x\)

\(x^2 y^2 = - \sin x\)

Answer 10

\(\cfrac{dy}{dx} = \cfrac{-(2xy^3 + y\cos x) }{3x^2 y^2 + \sin x}\) becomes :

\(\cfrac{dy}{dx} = \cfrac{-(2xy^3 + y\cos x) }{-3\sin x + \sin x} \\

\cfrac{dy}{dx} = \cfrac{-(2xy^3 + y\cos x) }{-2\sin x} \)

Answer 11

Or :

\(\cfrac{dy}{dx} = \cfrac{\cancel{-}(2xy^3 + y\cos x) }{\cancel{-}2\sin x}\)

Answer 12

\(\cfrac{dy}{dx} = \cfrac{2xy^3}{2\sin x} + \cfrac{y\cos x}{ 2\sin x}\)

\(\cfrac{dy}{dx} = {xy^3 \times \csc x} + \cfrac{y}{2} \cot x\)

Answer 13

That seems to be the finally answer, right? Or is there any simplification required?

Answer 14

Why are you changing x^2 y^2 to -sin x?

Answer 15

Because it is given in the question..

Answer 16

This first one is the final answer

Answer 17

Did you mean to say that the one without that simplificatin is the final answer?

Answer 18

*simplification..

Answer 19

\(\color{blue}{\text{Originally Posted by}}\) @mathslover
\(\cfrac{dy}{dx} = \cfrac{-(2xy^3 + y\cos x) }{3x^2 y^2 + \sin x}\)
\(\color{blue}{\text{End of Quote}}\)

This one?

Answer 20

Yes, I think so. I don't think your simplification is valid

Answer 21

Since 3x^2 y^2 is not equal to -3sin x

Answer 22

Oh, never mind!
I'm so sorry!

Answer 23

I see what you were doing now. My bad! >.<

Answer 24

Well, yeah, the condition stands TALL : \(y \ne 0\) ... if y = 0 I can not say that x^2y^2 = -sin x

Answer 25

see this is why not getting enough sleep does to you

Answer 26

Yep, that's all correct!

Answer 27

Although the answer will still work
You can check it manually
And if y = 0, then dy/dx = 0

Answer 28

Well yeah... why don't u have 2-3 days of continuou sleep... may be, this will help in future :P

Answer 29

But @Miracrown -> if y = 0

then I can NOT cancel out y in the equation x^2 y^3 + y sin x = 0

so, basically : y should not be zero for the final condition I got - \(\color{blue}{\text{Originally Posted by}}\) @mathslover
\(\cfrac{dy}{dx} = \cfrac{2xy^3}{2\sin x} + \cfrac{y\cos x}{ 2\sin x}\)

\(\cfrac{dy}{dx} = {xy^3 \times \csc x} + \cfrac{y}{2} \cot x\)
\(\color{blue}{\text{End of Quote}}\)

Answer 30

Unless x=0 too, and then who knows?

Answer 31

Yeah, you're right but there's a workaround
Take the very first equation you have -- the implicit one
And set y=0 in there

Answer 32

Yeah...!

So, if my final answer is \(\color{blue}{\text{Originally Posted by}}\) @mathslover
\(\cfrac{dy}{dx} = \cfrac{2xy^3}{2\sin x} + \cfrac{y\cos x}{ 2\sin x}\)

\(\cfrac{dy}{dx} = {xy^3 \times \csc x} + \cfrac{y}{2} \cot x\) ------ FINAL
\(\color{blue}{\text{End of Quote}}\)

Then , I will have to say that y should not be equal to zero. Right?

Answer 33

You get that dy/dx sin x = 0
So dy/dx = 0 unless sin x = 0
And in that case, it's undefined

Answer 34

So it works for that case even though the work didn't
You can say that EITHER y can't be 0 OR sin x can't be 0

Answer 35

dy/dx = 0 , then doesn't it mean that y = constant?

So , why is it undefined?

Answer 36

Out of curiosity, what if you didn't do implicit differentiation, divided out one of the y and solved the quadratic of y, then took the derivative? Does that mean there are now two derivatives and we're missing one some how?

Answer 37

y is only "constant" in very small areas

Answer 38

Well, we should get the same answer for the derivative
If we did it correctly

Answer 39

divide by y, and we get:
x^2 y^2 + sin x = 0
(Like before, we can only do this if y isn't 0)
Then x^2 y^2 = -sin x
So y^2 = -sin x/x^2
y = +/- sqrt(-sin x/x^2)

Answer 40

Clearly : \(y^2 = \cfrac{-\sin x}{x^2}\)

Answer 41

Yeah... mira got it..

Answer 42

Then we take the derivative of this
dy/dx = +/- 1/ sqrt(-sin x/x^2) * (-cos x x^2 + 2x sin x)/x^4

Answer 43

We have "-" inside the square roots, wouldn't that turn into something imaginary?

Answer 44

Oh.. sorry, got it.

Answer 45

And apparently that's the same thing

Answer 46

It would -- if sin x < 0
Good catch
But the same thing happens in the original equation
Look at the original equation -- if sin x < 0

Answer 47

Sorry -- all of this is a problem if sin x > 0
Then x^2 y^3 will have the same sign as y sin x
So their sum can't be 0
Which means it's undefined if sin x > 0

Basically, this is a weird problem

Answer 48

Oooh pretty. http://www.wolframalpha.com/input/?i=y%3Dsqrt%28-sinx%29%2Fx

Answer 49

Yeah, that looks cool!

Answer 50

But crazy too
You can see it's imaginary a lot of the time

Answer 51

\[\cfrac{dy}{dx} = \pm \cfrac{1}{\sqrt{ \left( \cfrac{-sin x}{x^2} \right) }} \times \cfrac{(-cos x \times x^2 + 2x sin x)}{x^4} \]

Answer 52

Is this what you meant @Miracrown ...

Answer 53

I must appreciate you all... You all have done well in confusing me :P

Answer 54

Actually if you graph it it'll look like this: