Shortcut for Second Derivative Using Product Rule

[Schrodinger's Dog]

Homework Statement

Have to use the product rule twice.

[tex]\frac{d}{dx}=x^4e^xsin(x)[/itex]<br /> <br /> I got about as far as the first use of the product rule then stalled when I had to use it again.<br /> <br /> got this:-<br /> <br /> $$(1+x^2)sin(x)+(x^4e^x)sin(x)$$<br /> <br /> but not this:-<br /> <br /> $$(4x^3e^x+x^4e^x)sin(x)+x^4e^xcos(x)$$<br /> <br /> Problem is the books explanation misses out the first step and proceeds to the second order differential and the answer.<br /> <br /> $$x^3e^x((4+x)sin(x)+xcos(x))$$<br /> <br /> Can't work out how to get there, any pointers? Or just show a complete working so I can see where I went wrong.<br /> <br /> Thanks in advance for any help.[/tex]

 

[daveb]

(fgh)' = f'gh + fg'h + fgh'. Assume f(x) = x^4, g(x) = sinx, h(x) = e^x. Then just factor out x^3*e^x.

 

[Schrodinger's Dog]

Interesting rule not seen it before at least in that form, I'll give it a whirl presently and let you know.

Ta.

 

[HallsofIvy]

It is a simple extension of the product rule
(fg)'= f'g+ fg'

(fgh)'= ((fg)h)'= (fg)'h+ fgh'= (f'g+ fg')h+ fgh'= f'gh+ fg'h+ fgh'.

The extension to the product of any number of functions should be obvious.

 

[Schrodinger's Dog]

Thanks a lot, yes, I'm sure this is explained in the text at some point, but haven't got there yet. Just got to differentiating mixed products and quotients section, no rules yet other than the usual, very useful though... I'm done for today but if I try your new rules and have an issue I'll get back to you tomorrow.

Appreciated

 

[cepheid]

The extension to the product of any number of functions should be obvious.

For some reason, I decided to try and write a general expression for the derivative of the product of 'n' functions, with the i^th function being denoted by f_i(x). I found it a bit tricky. This is the best I could come up with. What do you think?

$$\frac{d}{dx} \left( \prod_{i=1}^n f_i(x) \right) = \sum_{j =1}^n \left( \prod_{\substack{i = 1 \\ i \not= j}}^n f_i(x) \frac{df_j(x)}{dx} \right)$$

 

[VietDao29]

For some reason, I decided to try and write a general expression for the derivative of the product of 'n' functions, with the i^th function being denoted by f_i(x). I found it a bit tricky. This is the best I could come up with. What do you think?

$$\frac{d}{dx} \left( \prod_{i=1}^n f_i(x) \right) = \sum_{j =1}^n \left( \prod_{\substack{i = 1 \\ i \not= j}}^n f_i(x) \frac{df_j(x)}{dx} \right)$$

Yes, that's correct. Proof by Induction can do the trick nicely.

 

[Schrodinger's Dog]

Homework Statement

$\frac{x.tan(x)}{1+x^2}$

Another double differentiation I got stuck on we have to use the quotient then the product rule. I'm ok again with the first step but not sure how to use the product rule on a quotient ie $$\frac{f(x)}{g(x)}$$

I get $$\frac{(1+x^2)sec^2(x)-xtan(x)2x}{(1+x^2)^2}$$

What's thrown me is I don't know the differential of x.tan(x)

tan(x) is sec^2(x) according to the book, I derived that earlier with the quotient rule, I just realized where I went wrong here

I'll try and work out the differential, they don't want much do they a table of differentials would be nice at this point.

 

[cristo]

Ok, I deleted my last post as firstly, I guessed the question wrong, and secondly, LaTex was messing up! Let's try again:

Using the quotient rule, the derivative of

$$\frac{x.\tan(x)}{1+x^2}$$

is

$$\frac{(1+x^2)\frac{d}{dx}\left\{x\cdot tanx\right\}-x\cdot tanx \cdot 2x}{(1+x^2)^2}$$. Now, $\frac{d}{dx}(xtanx)=tanx+x\cdot sec^2 x$, hence we have that the derivative is:

$$\frac{tanx(1-x^2)+x sec^2x(1+x^2)}{(1+x^2)^2}$$

Now, you want to differentiate this. We can write this as two terms: $$\frac{tanx(1-x^2)}{(1+x^2)^2}+\frac{xsec^2x}{1+x^2}$$. Differentiate this term by term, using the quotient rule and the product rule. It's easiest if you write out with the d/dx's first when applying the quotient rule, then apply the product rule to the products.

 

[cristo]

I'll try and work out the differential, they don't want much do they a table of differentials would be nice at this point.

Try this wiki page. The trick to this sort of question is to go one step at a time, and use one rule at a time; otherwise, one just gets utterly confused!

 

[Schrodinger's Dog]

Yes I can see that now^^, although you've done it differently from the book, it looks less confusing than the way they did it, and although your answer differs from theirs, when I look at theirs I can see that your answer is the same, except without the partial fraction.

Thanks again.

Try this wiki page. The trick to this sort of question is to go one step at a time, and use one rule at a time; otherwise, one just gets utterly confused!

Grand I never thought of asking wiki saves me five minutes of effort. Actually I could of typed it into a maths prog, but it's better to do these things for yourself if you are not fluent like I.

Latex has gone kabluey for me too, what's displayed up there is not what is displayed in the source text at all, oh well nm.

 

[daveb]

For problems such as these, I like to generalize the problem by assigning f1 to the first function, f2 to the second, etc., solving that, then plugging in the expressions for the functions and finishing up. E.g., (fg/h)' = [h(f'g + g'f) - fgh']/h^2.

(I really need to learn latex!)

 

[Schrodinger's Dog]

Do you know I was really unaware that you could do this sort of thing and end up with very easy general ways of differentiating mixed products, as I said earlier although I think I might of deleted it by mistake; the first one by daveb and clarified by HOI, worked perfectly and gave the right answer without the need for laborious steps, keep 'em coming guys, and^ I'll try that one out tomorrow to see if it's a simpler way of doing it, which I've no doubt it is.

Hehe I have a formal written exam this year -unlike last year which was all assessment based ie much more numerous and complicated questions, that took days instead of hours to complete - and this stuff is certainly going to help, assuming they allow stuff that is not in the course into the exam.

I have already asked my tutor if this is allowable, we'll see, it isn't always as they want to see you do it the hard way, and a kind of maxim is you don't need anything outside of the course as such, and I certainly will keep doing that just so I know how to do it both ways.

AAMOI, I have now done both of these the hard way, and they both work out, it was somewhat a matter of looking here, and seeing different ways of simplifying the process and then adapting my method according to what I'd learned but mostly by doing it the other way and spotting fundamental errors of simplification etc. All very helpful and much appreciation.

 

[cristo]

Do you know I was really unaware that you could do this sort of thing and end up with very easy general ways of differentiating mixed products, as I said earlier although I think I might of deleted it by mistake; the first one by HOI, worked perfectly and gave the right answer without the need for laborious steps, keep 'em coming guys, and^ I'll try that one out tomorrow to see if it's a simpler way of doing it, which I've no doubt it is.

Hehe I have a written exam this year and this stuff is certainly going to help, assuming they allow stuff that is not in the course into the exam.

I have already asked my tutor if this is allowable, we'll see, it isn't always as they want to see you do it the hard way, and I certainly will keep doing that just so I know how to do it both ways.

AAMOI, I have now done both of these the hard way, and they both work out, it was mostly a matter of looking here, and seeing different ways of simplifying the process and then adapting there method according to what I'd learnt. All very helpful and much appreciation.

I think you should be able to use any technique you want-- after all, the course is presumably testing use of the differentiation rules, so simplifying things doesn't mean that you don't need to use the rules.

Still, it's always good, and time saving, to remember shortcuts to things in the future.

(I really need to learn latex!)

You don't even need to learn anything anymore-- click on the $\Sigma$ button at the top of the "reply to message" box

 

[Schrodinger's Dog]

Actually latex is very simple to learn I think, it took me 24 hours of messing around with different samples of the language to get a feel, and then practice made perfect. I think if you look at the source code of the latex on PF thread you can become proficient enough to get by in a day just by plagiarising other peoples code, and then after you use it for a while, you make less mistakes.

I agree cristo I can't really think of a reason they'd object so I'll be trying to memorise all the methods and using the easier one if I am allowed.

EDIT: My tutor said that double product method is totally acceptable, in fact he said whatever way works is fine

 

[Schrodinger's Dog]

For problems such as these, I like to generalize the problem by assigning f1 to the first function, f2 to the second, etc., solving that, then plugging in the expressions for the functions and finishing up. E.g., (fg/h)' = [h(f'g + g'f) - fgh']/h^2.

(I really need to learn latex!)

What is f(x)g(x)h(x)

I can't seem to get an answer with this one. Maybe I'm assigning the wrong values to the terms?

If you could work it through with your equation it would help.

 

[VietDao29]

What is f(x)g(x)h(x)

I can't seem to get an answer with this one. Maybe I'm assigning the wrong values to the terms?

If you could work it through with your equation it would help.

Well, I think f(x) = x, g(x) = tan(x), and h(x) = 1 + x²

So, the above expression can be re-written as:
$$\frac{x \tan x}{1 + x ^ 2} = \frac{f(x) g(x)}{h(x)}$$

By using The Quotient Rule, we have:
$$\left( \frac{f(x) g(x)}{h(x)} \right) '_x = \frac{\left[ f(x) g(x) \right]' h(x) - f(x) g(x) h'(x)}{h ^ 2 (x)} = \frac{h(x) \left[ f'(x) g(x) + f(x) g'(x) \right] - f(x) g(x) h'(x)}{h ^ 2 (x)}$$

You can just plug all the functions declared previously (i.e, f(x), g(x), and h(x)) in the final expression, and obtain the answer. Or you can do it your way, i.e step by step. They are pretty much the same. :)

 

[Schrodinger's Dog]

Ah ok thanks that's really great to see it outlined so distinctly, actually my problem was in missing out a single term which of course changes everything, I realized soon after(always check your answers thoroughly ), so it was just a brain fart . These are so neat and easy it's almost like cheating, though my tutor says there fine, as long as your working is detailed enough and answer is correct any method at all is acceptable.

 

[vishu]

what is latex rule.?
i m not aware about that...
pleasezzzz tell me is there any shortcut for second derivative