Need help evaluating results of implicit differentiation. « Next Oldest | Next Newest »

 ▼ Hal Bitton in Boise Senior Member Posts: 291 Threads: 43 Joined: Jun 2007 10-07-2010, 09:40 AM Hi everybody, After I do implicit differentiation on my 50G, and solve for d1y(x), I have my result, which is d1y(x) = expression in terms of x and y(x). So far so good. Now I want to plug constants in for x and y(x) and evaluate my derivative at a point. This is where I'm having trouble. I can't store a constant in the variable y(x), because that is apparently not a valid variable name. I tried using the define command to create a user function setting y(x) equal to a constant ..IE y(y(x))=3 , but that did not seem to work. Is there any way to tell the calc that y(x) is equal to 3 when evaluating the results of my implicit differentiation? Thanks, and best regards, Hal ▼ Crawl Senior Member Posts: 306 Threads: 3 Joined: Sep 2009 10-07-2010, 12:23 PM It seems like the easiest thing to do would be to go into Equation Editor, select Y(X) with the cursor, and replace it with 3. However, I can imagine situations where that wouldn't work (if there are a lot of Y(X)'s in the result and it would be tedious to replace them all by hand, or if you wanted to automate it with a program). So here's another way. When you have your expression, enter 'Y(X)' on the stack. Hit DERVX so you have d1Y(X) on the stack (or just enter 'd1Y(X)' to begin with). Then enter some other variable that you're not using, like Z. Then =, SUBST. Now you can enter 'Y(X)', 3, =, SUBST, and it'll substitute in, leaving the derivative alone (otherwise, it thinks you're saying Y(x) is the constant 3, so its derivative is 0). You just have to solve for Z now rather than d1Y(X). Or if you really want it back the other way, 'Z=d1Y(X)' SUBST at the end also works. ▼ Crawl Senior Member Posts: 306 Threads: 3 Joined: Sep 2009 10-07-2010, 12:43 PM Actually, if you've already solved for d1Y(X)=f(X,Y(X)), and you just want to evaluate the right side, you could just take the right side of the equation and do the 'Y(X)=3' SUBST step. (You can take the right side of an equation with OBJ->, DROP, DROP, SWAP, DROP) I was thinking about a more general situation, in which you either haven't solved for d1Y(X) yet, or for which it might be impossible to solve for it (except numerically) ▼ Hal Bitton in Boise Senior Member Posts: 291 Threads: 43 Joined: Jun 2007 10-08-2010, 01:17 PM Thanks very much. The SUBST command was exactly the function I was looking for. It works perfectly. Best regards, Hal

 Possibly Related Threads... Thread Author Replies Views Last Post HP Prime : inconsistency with implicit multiplication in CAS Olivier Lecluse 6 1,287 10-27-2013, 03:30 AM Last Post: parisse hp prime - sending program results to the stack giancarlo 6 1,150 10-15-2013, 02:00 AM Last Post: Giancarlo HP Prime complex results Javier Goizueta 0 566 10-06-2013, 12:59 PM Last Post: Javier Goizueta HP Prime Solving Nonlinear System of Equations for Complex Results Helge Gabert 11 2,255 09-30-2013, 03:44 AM Last Post: From Hong Kong 30B TVM slightly different results bill platt 5 1,078 03-26-2012, 11:21 AM Last Post: bill platt Complex results acceptance 32sII/33s/35s Matt Agajanian 9 1,398 03-24-2012, 09:30 AM Last Post: Dieter Wrong answer when evaluating log(10^-12) Daniel Greenidge 4 875 01-26-2012, 05:54 PM Last Post: Daniel Greenidge HP-70 erratic behaviour - first results Alberto Fenini 0 435 12-04-2011, 03:49 PM Last Post: Alberto Fenini HP15C LE Keyboard Survey Results M. Joury 23 2,760 10-02-2011, 07:52 PM Last Post: M. Joury Results of new root-seeking methods Namir 3 772 08-22-2011, 03:28 PM Last Post: C.Ret

Forum Jump: