Sorry to be a pain... but what's implicit differentiation?
- Thread starter Lizzie
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Thanks!
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The derivative of y is dy/dx. Implicit differentiation is, in a way, applying the Chain Rule to expressions in y. So the derivative, with respect to x, of y<sup>2</sup> is the derivative first of the square, times the derivative of the y.
. . . . .d(y<sup>2</sup>)/dx = 2y(dy/dx)
I'm surprised you're being asked questions on this topic before it's been covered in class. Would you like some links to online lessons on this topic?
Eliz.
. . . . .d(y<sup>2</sup>)/dx = 2y(dy/dx)
I'm surprised you're being asked questions on this topic before it's been covered in class. Would you like some links to online lessons on this topic?
Eliz.
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Sorry, but no. There are too many posts by too many students on too many forums at too many sites and... Okay, it's more likely that I'm just too old and forgetful...Lizzie said:
...um... what were we talking about...? :wink:
Online or otherwise, though, the topic should have been covered in class and/or in the text before you were assigned questions on the topic. To do otherwise is, at the least, fairly sloppy.
Sure thing!Lizzie said:
. . . . .Math Centre: Implicit Differentiation online tutorial
. . . . .Paul's Online Math Notes: Implicit Differentiation
. . . . .Implicit Differentiation (from MIT)
. . . . .UBC Calculus Online: Implicit Differentiation
. . . . .Wikipedia: Implicit Differentiation
. . . . .Math Help Central: Implicit Differentiation
Hope these help a bit.
Eliz.
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Glad they helped!
Just so you know: Implicit differentiation isn't usually that hard, but the problems can get extremely messy. Don't feel like you "must" have done something wrong if you find yourself in the middle of something extremely convoluted. On the "up" side, though, the word problems are generally extremely straightforward.
Eliz.
Just so you know: Implicit differentiation isn't usually that hard, but the problems can get extremely messy. Don't feel like you "must" have done something wrong if you find yourself in the middle of something extremely convoluted. On the "up" side, though, the word problems are generally extremely straightforward.
Eliz.
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If you'd had, say, "y = sin<sup>2</sup>(x)", you'd differentiate by doing the square first (two times whatever was inside the square), times the derivative of the sine (being cosine): dy/dx = 2sin(x)cos(x). And when you get down to "x", you're done, because dx/dx = 1, which you can safely ignore.
In the implicit-differentiation case, you don't have "y" by itself. You differentiate according to all the same rules, but when you get down to the variable, it isn't "x"; it's "y". And dy/dx does not equal "1", so you can't just ignore it. Instead, the last step, instead of "and now I'm down to 'x', so I'm done", is "and now I'm down to 'y', whose derivative is dy/dx".
When you're doing implicit-diff word problems, you'll mostly be differentiating with respect to time "t". So you'll have, say, boats moving at ninety-degree angles to each other (thus forming a right triangle), and you'll be asked to find how fast the distance between them is changing at some given time. You'll set up the distances according to the Pythagorean Theorem (which you'll use to death), getting x<sup>2</sup> + y<sup>2</sup> = h<sup>2</sup>. You'll be given the speeds of the boats, which you'll translate as dx/dt and dy/dt. You'll differentiate implicitly to get:
. . . . .2x(dx/dt) + 2y(dy/dt) = 2h(dh/dt)
You'll have been given the speeds dy/dt and dx/dt. You'll have figured out the values of x and y from the times and the speeds. You'll have found h from the Pythagorean Theorem. You'll plug in all the known values and solve for dh/dt.
Implicit diff is just like regular diff, except that you don't end with "x".
Eliz.
In the implicit-differentiation case, you don't have "y" by itself. You differentiate according to all the same rules, but when you get down to the variable, it isn't "x"; it's "y". And dy/dx does not equal "1", so you can't just ignore it. Instead, the last step, instead of "and now I'm down to 'x', so I'm done", is "and now I'm down to 'y', whose derivative is dy/dx".
When you're doing implicit-diff word problems, you'll mostly be differentiating with respect to time "t". So you'll have, say, boats moving at ninety-degree angles to each other (thus forming a right triangle), and you'll be asked to find how fast the distance between them is changing at some given time. You'll set up the distances according to the Pythagorean Theorem (which you'll use to death), getting x<sup>2</sup> + y<sup>2</sup> = h<sup>2</sup>. You'll be given the speeds of the boats, which you'll translate as dx/dt and dy/dt. You'll differentiate implicitly to get:
. . . . .2x(dx/dt) + 2y(dy/dt) = 2h(dh/dt)
You'll have been given the speeds dy/dt and dx/dt. You'll have figured out the values of x and y from the times and the speeds. You'll have found h from the Pythagorean Theorem. You'll plug in all the known values and solve for dh/dt.
Implicit diff is just like regular diff, except that you don't end with "x".
Eliz.
Thanks, I found out why I was getting confused. The problem was a multiple choice one and none of the answers were correct, so i was getting confused. But, I then noticed that there was another choice I had obviously overlooked which said e- none of the above, lol. Thanks.