PDA

View Full Version : Quick Calc Question

ebony
09-06-2006, 07:56 PM
What is d/dx of y^x ?

Thanks all

Kazodev
09-06-2006, 08:00 PM
What is d/dx of y^x ?

Thanks all

that doesn't really make sense, you differentiate equations i.e. something with an "=" sign. For example you don't really differentiate x^3, you differentiate both sides of y = x^3. Now d/dx of y is dy/dx and d/dx of x^3 is 3x^2, thus we say dy/dx = y' = 3x^2. Now assuming you have y = y^x, then there are a few ways to deal with this, one way is to take ln of both sides, thus we get ln|y| = x*ln|y|, differentiating both sides we get y'/y = ln|y| + x*y'/y and now you can solve for y. Another way, assuming once again you have y = y^x, is to note that IF y <> 0, then 1 = y^(x-1) and then do the ln trick again or whatever way you choose. Is there a more complete question you can post since I don't think you have y = y^x. If you had for example z = f(x,y) = y^x, that turns it into a completely different problem...

wat?
09-06-2006, 08:00 PM
Assuming y is a constant, d/dx of y^x = (ln y) * y^x

G.V.Ramanathan
09-07-2006, 09:59 AM
What is d/dx of y^x ?

Thanks all

that doesn't really make sense, you differentiate equations i.e. something with an "=" sign. For example you don't really differentiate x^3, you differentiate both sides of y = x^3. Now d/dx of y is dy/dx and d/dx of x^3 is 3x^2, thus we say dy/dx = y' = 3x^2. Now assuming you have y = y^x, then there are a few ways to deal with this, one way is to take ln of both sides, thus we get ln|y| = x*ln|y|, differentiating both sides we get y'/y = ln|y| + x*y'/y and now you can solve for y. Another way, assuming once again you have y = y^x, is to note that IF y <> 0, then 1 = y^(x-1) and then do the ln trick again or whatever way you choose. Is there a more complete question you can post since I don't think you have y = y^x. If you had for example z = f(x,y) = y^x, that turns it into a completely different problem...

I am afraid it is this post that does not make any sense to me. And it is very misleading. How do you differentiate an equation? How do you define the derivative of an equation? 4*2 = 8 is an equation. What is its derivative?

It is functions that you differentiate. Yes, you differentiate x^3 and its derivative is 3 x^2.
(Note that this post also says that (d/dx) x^3 = 3x^2. !)

Transforming the “quick question” to something entirely different and then taking the logarithm of both sides and then differentiating and then “solving” what you get leads to utter nonsense. The relation y = y^x is an identity if x = 1 and if x is not equal to 1 then the relation has one or two solutions, namely y = 0 or y = 1, or both.

Ebony: I should advice you to take wat?’s clear, crisp and correct answer to your “quick question.”

(wat?’s answer can be got by noting that (y^x)’ = [exp(x ln y) ]’
= exp(x ln y) ln y (by chain rule) = y^x ln y.)

If any one asks you why you did not say that z = y^x is a function of two variables, then tell them it was not easy to post partial derivative symbols, and that wat?’s answer does not change.

If any one asks why you did not explicitly say y > 0, ask them to take a course in complex analysis and come back.

Kazodev
09-07-2006, 11:23 AM
that doesn't really make sense, you differentiate equations i.e. something with an "=" sign. For example you don't really differentiate x^3, you differentiate both sides of y = x^3. Now d/dx of y is dy/dx and d/dx of x^3 is 3x^2, thus we say dy/dx = y' = 3x^2. Now assuming you have y = y^x, then there are a few ways to deal with this, one way is to take ln of both sides, thus we get ln|y| = x*ln|y|, differentiating both sides we get y'/y = ln|y| + x*y'/y and now you can solve for y. Another way, assuming once again you have y = y^x, is to note that IF y <> 0, then 1 = y^(x-1) and then do the ln trick again or whatever way you choose. Is there a more complete question you can post since I don't think you have y = y^x. If you had for example z = f(x,y) = y^x, that turns it into a completely different problem...

I am afraid it is this post that does not make any sense to me. And it is very misleading. How do you differentiate an equation? How do you define the derivative of an equation? 4*2 = 8 is an equation. What is its derivative?

It is functions that you differentiate. Yes, you differentiate x^3 and its derivative is 3 x^2.
(Note that this post also says that (d/dx) x^3 = 3x^2. !)

Transforming the “quick question” to something entirely different and then taking the logarithm of both sides and then differentiating and then “solving” what you get leads to utter nonsense. The relation y = y^x is an identity if x = 1 and if x is not equal to 1 then the relation has one or two solutions, namely y = 0 or y = 1, or both.

Ebony: I should advice you to take wat?’s clear, crisp and correct answer to your “quick question.”

(wat?’s answer can be got by noting that (y^x)’ = [exp(x ln y) ]’
= exp(x ln y) ln y (by chain rule) = y^x ln y.)

If any one asks you why you did not say that z = y^x is a function of two variables, then tell them it was not easy to post partial derivative symbols, and that wat?’s answer does not change.

If any one asks why you did not explicitly say y > 0, ask them to take a course in complex analysis and come back.
I agree that you can only take derivatives of functions but what if we have x^2 + y^2 = r^2, for some r > 0. That's not a function yet the question can ask for dy/dx? It wouldn't require easy or difficulty to post partial derivative symbols, if I said z = f(x,y) = y^x, then depending on whether I ask df/dx or df/dy would mak it clear which partial they're refering to or a simple f_x(x,y) vs f_y(x,y). Still seems like an odd question to post in course 2/FM forum if the question was about partial derivatives, I didn't encounter too many of them on FM.

ACCtuary
09-07-2006, 11:42 AM
Guys, this is implicit differentiation.

If x^2+y^2=r^2, differentiate both sides with respect to x
2x + 2y (dy/dx) = 0

The second term is differentiated via the chain rule, assuming that y is (locally) a function of x around some point (x_0,y_0).

Thus we have (dy/dx)=-x/y which is the same answer you'd get if you solved for y and then differentiated. Note that this result is only valid if y is not 0, which makes sense since the circle has a vertical tangent at these points.

atomic
09-07-2006, 11:46 AM
What if we have x^2 + y^2 = r^2, for some r > 0. That's not a function yet the question can ask for dy/dx? It wouldn't require easy or difficulty to post partial derivative symbols, if I said z = f(x,y) = y^x, then depending on whether I ask df/dx or df/dy would mak it clear which partial they're refering to or a simple f_x(x,y) vs f_y(x,y). Still seems like an odd question to post in course 2/FM forum if the question was about partial derivatives, I didn't encounter too many of them on FM.

It's still a mapping, and a derivative can be computed implicitly--that is to say, the derivative of one variable with respect to another is done by treating the dependent variable as an implicit function of the independent variable. Derivatives are operators on maps, and certain conditions must apply on the map in order for the derivative to be well-defined. But how useful is this information to someone who just wants to know the derivative of y^x with respect to x?

In the OP's original question, the mapping is simply x \mapsto y^x. No domain or codomain is specified, nor is a relationship between y and x assumed. If you really wanted to show your knowledge, you would have asked whether y is an implicit function of x, in which case the derivative would be

\frac{d}{dx}\left[y^x\right] = \frac{d}{dx}\left[e^{x \log y}\right] = \frac{d}{dx}[x \log y] e^{x \log y} = \left(\log y + \frac{x}{y} \frac{dy}{dx}\right)y^x.

Nothing more needs to be done. As Ramanathan pointed out, it is incorrect to then equate dy/dx with the derivative just computed. Note that the above result simplifies correctly when y is independent of x; i.e., dy/dx = 0.

Let's try not to get too pedantic about someone's rather simple question, and instead of falling over oneself in an attempt to be clever, just make some simple (and logical, given the actuarial context) assumptions as wat? did.

Kazodev
09-07-2006, 11:47 AM
Guys, this is implicit differentiation.

If x^2+y^2=r^2, differentiate both sides with respect to x
2x + 2y (dy/dx) = 0

The second term is differentiated via the chain rule, assuming that y is (locally) a function of x around some point (x_0,y_0).

Thus we have (dy/dx)=-x/y which is the same answer you'd get if you solved for y and then differentiated. Note that this result is only valid if y is not 0, which makes sense since the circle has a vertical tangent at these points.

Exactly but the point was x^2 + y^2 = r^2 is not a function, it can be split up into 2 functions I guess and is that what you mean by locally?

Kazodev
09-07-2006, 11:48 AM
What if we have x^2 + y^2 = r^2, for some r > 0. That's not a function yet the question can ask for dy/dx? It wouldn't require easy or difficulty to post partial derivative symbols, if I said z = f(x,y) = y^x, then depending on whether I ask df/dx or df/dy would mak it clear which partial they're refering to or a simple f_x(x,y) vs f_y(x,y). Still seems like an odd question to post in course 2/FM forum if the question was about partial derivatives, I didn't encounter too many of them on FM.

It's still a mapping, and a derivative can be computed implicitly--that is to say, the derivative of one variable with respect to another is done by treating the dependent variable as an implicit function of the independent variable. Derivatives are operators on maps, and certain conditions must apply on the map in order for the derivative to be well-defined. But how useful is this information to someone who just wants to know the derivative of y^x with respect to x?

If you really wanted to show your knowledge, you would have asked whether y is an implicit function of x, in which case the derivative would be

\frac{d}{dx}\left[y^x\right] = \frac{d}{dx}\left[e^{x \log y}\right] = \frac{d}{dx}[x \log y] e^{x \log y} = \left(\log y + \frac{x}{y} \frac{dy}{dx}\right)y^x.

Nothing more needs to be done. As Ramanathan pointed out, it is incorrect to then equate dy/dx with the derivative just computed. Note that the above result simplifies correctly when y is independent of x; i.e., dy/dx = 0.

Let's try not to get too pedantic about someone's rather simple question, and instead of falling over oneself in an attempt to be clever, just make some simple (and logical, given the actuarial context) assumptions as wat? did.
I don't believe I was falling over myself to be clever, I asked for clarification and wrote what I'd do based on what I assumed.

G.V.Ramanathan
09-07-2006, 11:53 AM
x^2 + y^2 = r^2 DOES define y as a function of x. dy/dx means the derivative of the implicit function as defined by that equation. You differentiate the function on the left-hand side and the function on the right-hand side and set them to be equal.

If z = f(x,y) = y^x, then originally posted question, which asks for (d/dx) y^x, makes it abundantly clear that the derivative is with respect to x.

What one should NOT do is set y = y^x and then attempt to take logarithm and differentiate. That is a blunder which should not be propagated.

This is not a complicated question. It has a simple and direct answer as wat? has posted.

ACCtuary
09-07-2006, 11:55 AM
The answer to the question as originally posted (using differential notation)

\exp(x\log y) d(x\log y)

which equals

\exp(x\log y) ((\log y) dx + d(\log y)*x)

Thus the differential of this expression is
\exp(x\log y) * ((\log y) dx + x/y dy)

And if we assume y is also a function of x, the derivative at a point can be calculated by the formula

y^x * (\log y + (x/y) (dy/dx)

G.V.Ramanathan
09-07-2006, 11:58 AM
I posted without realizing that Atomic has already explained it very well! I apologize for the repetition.

ACCtuary
09-07-2006, 11:59 AM
Guys, this is implicit differentiation.

If x^2+y^2=r^2, differentiate both sides with respect to x
2x + 2y (dy/dx) = 0

The second term is differentiated via the chain rule, assuming that y is (locally) a function of x around some point (x_0,y_0).

Thus we have (dy/dx)=-x/y which is the same answer you'd get if you solved for y and then differentiated. Note that this result is only valid if y is not 0, which makes sense since the circle has a vertical tangent at these points.

Exactly but the point was x^2 + y^2 = r^2 is not a function, it can be split up into 2 functions I guess and is that what you mean by locally?

Locally:

Near a particular point. If i say find the slope of the tangent to the circle x^2+y^2 = 1 near (3/5,4/5), you would use the formula m = - x/y = -3/4.
Of couse, this is just what we know from high school geometry - that the tangent is perpendicular to the radius.

Y is a function of x near (3/5,4/5). Just solve for y in terms of x. y=\sqrt{1-x^2}. But the fun of implicit differentiation is you can play this game even if doing the algebra is ugly (or impossible). You just have to verify the curve won't have a vertical tangent.

For instance, if x^2+xy-y^2 = 19 near the point (4,3) you can calculate the derivative without solving for y in terms of x. Even though you don't carry out the algebra, you assume y is some function of x and go ahead and differentiate both sides.

The details are left as an exercise for the reader, as is identifying the conic section mentioned herein. (The second question is not important for the exam!). Once you have a formula for the first derivative, you can play the game again to get the second derivative, and so on.

Kazodev
09-07-2006, 12:02 PM
x^2 + y^2 = r^2 DOES define y as a function of x.
I thought function required that x cannot map to 2 different y's, clearly here (0,r) and (0,-r) work but r != -r.

What one should NOT do is set y = y^x and then attempt to take logarithm and differentiate. That is a blunder which should not be propagated.

This is not a complicated question. It has a simple and direct answer as wat? has posted.

Yes my error lead to nonsense, I agree there

I still don't get why a question on partial derivatives was asked in course 2/FM forum. Seems like even 1/P would've been closer to the topic or General

ACCtuary
09-07-2006, 12:07 PM
I posted without realizing that Atomic has already explained it very well! I apologize for the repetition.

As did I.

G.V.Ramanathan
09-07-2006, 12:10 PM
I thought function required that x cannot map to 2 different y's, clearly here (0,r) and (0,-r) work but r != -r.

Please see the clause that follows. It is understood that the implicit function is defined locally. There are posts above that describe what is meant by "locally. "

ACCtuary
09-07-2006, 12:10 PM
I thought function required that x cannot map to 2 different y's, clearly here (0,r) and (0,-r)

The key word here is locally . If you fix a point, you can choose only one half of the semi-circle. That half will define y as a function of x.

Thus continuing the example before, (3/5,4/5) is on the upper half and you would use y=sqrt(1-x^2) to differentiate. (0,-r) is on the lower half, you would use y=-sqrt(1-x^2) to differentiate.

Of course, actually solving is too tedious. Just differentiate both sides, remembering to use the chain rule for functions of y, and you should be fine.

What makes this game work is that you are only looking at a neighborhood of a given point. Near that point, the function passes the "vertical line test"

I need a good picture here. I'll post it later.

Kazodev
09-07-2006, 12:21 PM
I thought function required that x cannot map to 2 different y's, clearly here (0,r) and (0,-r)

The key word here is locally . If you fix a point, you can choose only one half of the semi-circle. That half will define y as a function of x.

Thus continuing the example before, (3/5,4/5) is on the upper half and you would use y=sqrt(1-x^2) to differentiate. (0,-r) is on the lower half, you would use y=-sqrt(1-x^2) to differentiate.

Of course, actually solving is too tedious. Just differentiate both sides, remembering to use the chain rule for functions of y, and you should be fine.

What makes this game work is that you are only looking at a neighborhood of a given point. Near that point, the function passes the "vertical line test"

I need a good picture here. I'll post it later.
No it's fine, I can visualize the picture. Locally was just the key word I was looking for. Thanks acc and atomic. However my other question remains unanswered, ebony?

ebony
09-08-2006, 03:34 PM
Wow! I have been away for a few days and I think I mistakenly I lead you all down a path of discussion that wasn't intended by my original question. Sorry about that:oops: .

My original question was to help me in finding the answer to:
d/dd (1-v^10)/d = -33.865

(yes this is meant to be a continuous annuity. See below.)

My question was meant to focus on how to take the derivative of the v^10 portion of this problem once it has been transformed to e^(-10d).

Wat - you read my mind and helped me with what I was looking for.

BTW: How are you able to type equations into your posts? They are much easier to follow that way!

wat?
09-08-2006, 03:35 PM
Wat - you read my mind and helped me with what I was looking for.

No problem. :)

Kazodev
09-08-2006, 03:38 PM
BTW: How are you able to type equations into your posts? They are much easier to follow that way!