Why implicit differentiation?

Hi MHF. I have a short question with hopefully a simple answer.

Why do we use implicit differentiation? What is the difference between a problem where we don't need implicit differentiation and a problem where we do need it?

It seems to me that you can solve every problem explicitly, but maybe I just haven't gone far enough or perhaps later the problems become unbearable while computing them explicitly?

Thanks!

Re: Why implicit differentiation?

you cannot solve every problem explicity, for example compute $\displaystyle \frac{dy}{dx} $ (for points which are defined and meets criteria for implicit function theorem to be valid at that point) of $\displaystyle y^2 + y = x $, you cannot isolate this in the form y = f(x).

Re: Why implicit differentiation?

Quote:

Originally Posted by

**jakncoke** you cannot solve every problem explicity, for example compute $\displaystyle \frac{dy}{dx} $ (for points which are defined and meets criteria for implicit function theorem to be valid at that point) of $\displaystyle y^2 + y = x $, you cannot isolate this in the form y = f(x).

Hmm.

I see.

So in your problem we have $\displaystyle y'=\frac {1}{1+2y}$. So the slope at any point on the function is $\displaystyle \frac{1}{1+2y}$ correct?

Because I know the derivative of your function. But if you ask me to differentiate it at $\displaystyle x=3$ can I just plug in $\displaystyle y^2+y=3$ and differentiate to find the slope at x=3?

Thanks!

Re: Why implicit differentiation?

Quote:

Originally Posted by

**jakncoke** you cannot solve every problem explicity, for example compute $\displaystyle \frac{dy}{dx} $ (for points which are defined and meets criteria for implicit function theorem to be valid at that point) of $\displaystyle y^2 + y = x $, you cannot isolate this in the form y = f(x).

Actually, in this case you can.

$\displaystyle \displaystyle \begin{align*} y^2 + y &= x \\ y^2 + y + \left( \frac{1}{2} \right)^2 &= x + \left( \frac{1}{2} \right)^2 \\ \left( y + \frac{1}{2} \right)^2 &= x + \frac{1}{4} \\ \left( y + \frac{1}{2} \right)^2 &= \frac{4x + 1}{4} \\ y + \frac{1}{2} &= \pm \frac{\sqrt{ 4x + 1 }}{2} \\ y &= \frac{ -1 \pm \sqrt{ 4x + 1 } }{2} \end{align*}$

But this demonstrates the point that while it may be possible to get an explicit form of y = f(x), it would be much more DIFFICULT to try to differentiate this explicit form, than it would be to differentiate the simple polynomial that it started with.

Another good example is to find the derivative of the circle $\displaystyle \displaystyle \begin{align*} x^2 + y^2 &= 1 \end{align*}$. To do this explicitly, we would need to write $\displaystyle \displaystyle \begin{align*} y = \pm \sqrt{ 1 - x^2 } \end{align*}$, which is a much more difficult expression to differentiate than the simple polynomials that we have started with.

Re: Why implicit differentiation?

Prove it showed that my example was flawed but in doing so he also illustrated why somtimes its better to use implicit differentiation even if you can write the equation explicity. Also in order to to quell your doubts as to if any equations exist which cannot be made explicit y = f(x), The example given on wikipedia is $\displaystyle y^5 - y = x $ (I'm a bit tired to verify if it is correct so take it at face value or verify it thy self)

Re: Why implicit differentiation?

Quote:

Originally Posted by

**jakncoke** Prove it showed that my example was flawed but in doing so he also illustrated why somtimes its better to use implicit differentiation even if you can write the equation explicity. Also in order to to quell your doubts as to if any equations exist which cannot be made explicit y = f(x), The example given on wikipedia is $\displaystyle y^5 - y = x $ (I'm a bit tired to verify if it is correct so take it at face value or verify it thy self)

Yea, I noticed that too when I looked deeper.

Can you answer me question about simply plugging in x=3 to solve for 3...Would that give me the slope at x=3?

Re: Why implicit differentiation?

Quote:

Originally Posted by

**Paze** Can you answer me question about simply plugging in x=3 to solve for 3...Would that give me the slope at x=3?

No, when you set x to 3, you have an equation with y only, so you no longer have a function to take the derivative of.

- Hollywood

Re: Why implicit differentiation?

Quote:

Originally Posted by

**hollywood** No, when you set x to 3, you have an equation with y only, so you no longer have a function to take the derivative of.

- Hollywood

Ow...How would I go about finding the slope at x=3 then (or any valid x value)? Do I need to know the y value...That sounds counter-productive since I suppose I need a x value to get a y value. I'm confused (Doh)

Re: Why implicit differentiation?

You would need to find the derivative function first, then assign x=3. And if you do implicit differentiation and end up with the derivative as a function of x and y, you'll need to figure out y and plug that in, too.

- Hollywood