<p>Yeah. I am doing calc right now. I am having trouble understanding the concept of respecting variables:</p>
<p>the whole </p>
<p>dx
-- <- fraction thingy
dy</p>
<p>and implicit differentiation.</p>
<p>Could someone explain it, and maybe give some examples! thanks in advance!</p>
<p>What do you need explained about it? Implicit differentiation is when the function cannot unambiguously define y in terms of x. So you might have y^2 + xy = 2 and you need to differentiate. The technique to d/dx'ing the y's is the same as the x's, the only trick is that after you differentiate you multiply by dy/dx. So for example:</p>
<p>y^3 + x^2<em>y + 3 = 0
This would differentiate to:
3y^2</em>dy/dx + 2xy + x^2*dy/dx = 0</p>
<p>OK. So basically, it is for situations where it is very difficult to get y by itself on one side. How come the books shows an example with:</p>
<p>x^2 + y^2 = 9</p>
<p>couldn't you just do y = (+/-)sqrt(9 - x^2) and differentiate that?</p>
<p>but y = (+/-) sqrt( ... ) isn't the same as x^2 + y^2 = 9 right?</p>
<p>You can only differentiate one function at a time (for practical purposes here...) so you can only differentiate y = + sqrt( ... ) or y = -sqrt( ... ). You need the entire circle at once, which is where implicit differentiation comes in.</p>
<p>Ok. So basically, you can't differentiate both at the same time. Could you explain about the whole multiplying both sides by dy/dx? I still don't get it... why do we multiply the y?</p>
<p>Your standard derivative -- i.e., the one you have been doing before you learned about implicit differentiation [aka explicit differentiation] is the same thing as differentiating with respect to x.</p>
<p>Example :</p>
<p>y = 5x^3 - 7x
y' = 15x^2 - 7</p>
<p>is the same thing as</p>
<p>y = 5x^3 - 7x
d/dx (y) = d/dx (5x^3 - 7x)
dy/dx = 15x^s - 7.</p>
<p>Think of it like this. When you take the derivative with respect to x, that's taking d/dx of each sides of the equation. The d/dx of y becomes dy/dx, or the derivative of y with respect to x. On the other hand, the d/dx of x, becomes dx/dx, which is 1 and is immaterial. In other words, the change in x divided by the change of x is 1 even as it goes to zero.</p>
<p>It's the same idea for implicit differentiation. The only difference is that after differentiation, you will have an equation implicating dy/dx as opposed to having dy/dx directly equal to a given expression.</p>
<p>Ok. Thanks SDFried. So after you multiply both sides by (dy/dx) you solve for (dy/dx) and so you can then proceed as you do normally.</p>
<p>Whenever you take the derivative of a term, you must multiply it by dx, dy, dz, etc., to indicate that you have taken the derivative of that variable. (this is necessary for integrals later on)</p>
<p>for example the derivative of y would be (1)dy and because we are doing it with respect to x, it becomes dy/dx</p>
<p>the derivative of x would be (1)dx and because it is with respect to x, it is dx/dx, thus 1</p>
<p>if you have y = x^2, then you take the derivative of both sides (not just multiply by dy/dx as you said in your last post) and end up with:
the derivative of y with respect to x equal to the derivative of x^2 with respect to x:
(1)(dy)/dx = (2x)(dx)/dx
dy/dx = 2x</p>
<p>for implicit differentiation (when you cannot neatly seperate the y's from the x's), you take the derivative of the entire function with respect to x.</p>
<p>y^3 + xy +3 = 0<br>
(for the xy you would do product rule)
(3y^2)(dy)/dx + x(1)(dy)/dx + y(1)(dx)/dx + 0/dx = 0/dx
(3y^2)(dy/dx) + x(dy/dx) + y = 0
then you seperate out all the dy/dx and put them on one side of the equation</p>
<p>(dy/dx) = -y/(3y^2 + x)</p>
<p>hope this helps. it is important for more advanced calculus that you know the basic rules i.e. when you take the derivative of y it is (1)dy and x is (1)dx. once you are comfortable with it all, by all means think ahead and skip steps, but know where your answers come from. Otherwise when you get into integrals, U-substitution and integration by parts wont make much sense.</p>
<p>nice explanation gospurs50 :)</p>
<p>Another way of thinking about it is the chain rule (I hope you've covered this already).</p>
<p>But, if you have y^2 + x^2 = 25, and you want to differentiate the whole function, you get: 2y<em>dy/dx + 2x</em> dx/dx = 0.
dx/dx = 1, so you're left with
2y*dy/dx + 2x = 0</p>
<p>Remember, the chain rule says when given a composite function (which, confusingly enough is technically all functions), you must take the derivative of the whole thing, ignoring the "inner-function", and multiply that by the derivative of just the "inner-function".</p>
<p>Hope this helps :)</p>