My Math Forum (http://mymathforum.com/math-forums.php)
-   Calculus (http://mymathforum.com/calculus/)
-   -   Are partial derivatives the same as implicit derivatives? (http://mymathforum.com/calculus/39808-partial-derivatives-same-implicit-derivatives.html)

 johngalt47 November 22nd, 2013 04:21 PM

Are partial derivatives the same as implicit derivatives?

I am about to pull my hair out. Some teachers and books refer to partial derivatives while others refer to implicit differentiation. Are they the same thing? Please explain.

 HallsofIvy November 22nd, 2013 05:06 PM

Re: Are partial derivatives the same as implicit derivatives

No, they are not at all the same thing. "Some texts and teachers" refer to "implicit differentiation" when they are referring to , the function y as given by an implicit function of x. That is, we are given something like f(x, y)= constant. We could, at least theoretically, solve that equation for y [b]as a function of x. In either case we have y as a function of the single variable, x.

We take partial derivatives of a function of more than one variable. That is, we do NOT have some function of x and y equal to a constant, it is the function of x and y (and z and u and whatever other independent variables we need), it is the function, f, itself we differentiate.

What may be confusing you is that, to find an implicit differentiation, we can use a partial derivative. To take a very simple example, suppose we are given y, defined as a function of x, by f(x,y)= 2x+ y= 7. We could immediately solve that equation for y: y= 7- 2x. Differentiating y with respect to x, dy/dx= -2.

Or we could use the "chain rule" to say that, with f= 2x+ y= 7 a function of x and y and y a function of x, $\dfrac{df}{dx}= \dfrac{\partial f}{\partial x}+ \dfrac{\partial f}{\partial y}\dfrac{dy}{dx}$. Since f(x,y)= 2x+ y, [latex]\dfrac{\partial f}{\partial x}= 2[latex] and $\dfrac{\partial f}{\partial y}= 1$. But f(x,y)= 2x+y= 7 also means that f(x,y) is the constant, 7, so its derivative, with respect to x, is 0.

That is, "$\dfrac{df}{dx}= \dfrac{\partial f}{\partial x}+ \dfrac{\partial f}{\partial y}\dfrac{dy}{dx}$" becomes  which we can solve to get, as before, $\dfrac{dy}{dx}= -2$.

So, while we can use "partial derivatives" to find the "implicit derivative", no, they are NOT the "same thing".

 johngalt47 November 23rd, 2013 11:44 AM

Re: Are partial derivatives the same as implicit derivatives

Ok, so what do you you to find the point at which a plane is tangent to a sphere?

 All times are GMT -8. The time now is 11:24 PM.