1. ## Differentiating y

I can't seem to wrap my head around differentiating y or f(x). For example: x^2+xy^2=2.

I understand the product rule and everyone keeps saying that ''it's just the product rule'' when I'm differentiating implicitly like that, but I don't have the intuition, it's just missing. Can someone attempt to clarify please? You do not need to use my example.

2. ## Re: Differentiating y

Can you show how you would use implicit differentiation on your example equation? It would help us to better help you if we see where exactly you are stuck.

3. ## Re: Differentiating y

$\displaystyle x^2+xy^2=2$

Which goes to,
$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^2+ \frac{\mathrm{d} }{\mathrm{d} x} xy^2 = \frac{\mathrm{d} }{\mathrm{d} x}2$

Now, you diffentiate any 'x' variable as per normal, but when you are differentiating with respect to 'y', you leave the differential operator there, I will do so in the '$\displaystyle y'$' form

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^2+ \frac{\mathrm{d} }{\mathrm{d} x} xy^2 = \frac{\mathrm{d} }{\mathrm{d} x}2$

I will do this in 3 parts.
The first term $\displaystyle x^2$
$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^2 = 2x$

The second term $\displaystyle xy^2$
$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} xy^2 = x\cdot2y\cdot y' + 1\cdot y^2$

The third term $\displaystyle 2$
$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}2 = 0$

All up, we get:
$\displaystyle 2x + 2xyy' + y^2 = 0$

$\displaystyle 2xyy' = -(y^2 + 2x)$

$\displaystyle y' = -\frac{y^2 + 2x}{2xy}$

4. ## Re: Differentiating y

Thanks, I realize how to actually perform the calculation but I'm not interested in plugging and playing so to speak. I'm trying to understand exactly why we 'leave the differential operator there'. It has something to do with y being a function and the product rule but my brain is not making the complete connection.

5. ## Re: Differentiating y

Basically what I'm trying to understand is this sentence: The derivative of y is y’. The derivative of x is taken in the normal way of a variable, so that the derivative of x is 1, as an example. The reason for this is that y is a function of x, and does not exist without being dependent on the equation on the other side, on the other hand, x exists all along the domain. I mentioned this briefly in my overview of derivatives.

6. ## Re: Differentiating y

Take this equation for example.
$\displaystyle x^2 + yx = 2$

It isn't explicitly a function of y, with respect to x, you have to do some manipulation before differentiating normally..

$\displaystyle y = ?$

In this case,

$\displaystyle y=\frac{2 - x^2}{x}$

If I gave you this originally, you would be able to differentiate this, and the problem is no different.

But, one thing that we do usually forget is that we differentiate both sides.

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}y=\frac{\mathrm{d}y }{\mathrm{d} x}\frac{2 - x^2}{x}$

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}y=\frac{\mathrm{d}y }{\mathrm{d} x}2x^{-1}-\frac{\mathrm{d}y}{\mathrm{d} x}x$

So we get,
$\displaystyle 1\cdot \frac{\mathrm{d}y }{\mathrm{d} x}=-2x^{-2}-1$

Take note of the left hand side.

Finally we would get:
$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}=-2x^{-2}-1$

7. ## Re: Differentiating y

Take this equation for example.
$\displaystyle x^2 + yx = 2$

It isn't explicitly a function of y, with respect to x, you have to do some manipulation before differentiating normally..

$\displaystyle y = ?$

In this case,

$\displaystyle y=\frac{2 - x^2}{x}$

If I gave you this originally, you would be able to differentiate this, and the problem is no different.

But, one thing that we do usually forget is that we differentiate both sides.

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}y=\frac{\mathrm{d}y }{\mathrm{d} x}\frac{2 - x^2}{x}$

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}y=\frac{\mathrm{d}y }{\mathrm{d} x}2x^{-1}-\frac{\mathrm{d}y}{\mathrm{d} x}x$

So we get,
$\displaystyle 1\cdot \frac{\mathrm{d}y }{\mathrm{d} x}=-2x^{-2}-1$

Take note of the left hand side.

Finally we would get:
$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}=-2x^{-2}-1$
so if the equation was instead:

$\displaystyle y^2=\frac{2 - x^2}{x}$

$\displaystyle 2\cdot \frac{\mathrm{d}y }{\mathrm{d} x}=-2x^{-2}-1$ ?

8. ## Re: Differentiating y

$\displaystyle 2y\cdot \frac{\mathrm{d} y}{\mathrm{d} x}= -2x^{-2}-1$

If it were,
$\displaystyle y^{3} = ????$

The derivative would be,
$\displaystyle 3y^{2}\cdot \frac{\mathrm{d} y}{\mathrm{d} x}= -2x^{-2}-1$

If it were,
$\displaystyle y^{4} = ????$

The derivative would be,
$\displaystyle 4y^{3}\cdot \frac{\mathrm{d} y}{\mathrm{d} x}= -2x^{-2}-1$

If it were,
$\displaystyle e^{y} = ????$

The derivative would be,
$\displaystyle e^{y}\cdot \frac{\mathrm{d} y}{\mathrm{d} x}= -2x^{-2}-1$

etc...