Derivative of a Series: How to Solve for x^n/((x-a1)(x-a2)...(x-an)) in Calculus

[Physics lover]

Homework Statement

if y=1+a1/(x-1) +a2x((x-a1)(x-a2)) +a3x^2/((x-a1)(x-a2)(x-a3))+....+anx^n-1/((x-a1)(x-a2)...(x-an)).
Find dy/dx.

Sorry i haven't used LaTeX this time as i will first have to learn it.

Relevant Equations

Logarithmic differentiation

I first solved the first two terms and then i solved the resulting term with the third term and so on.At last i was left with x^n/((x-a1)(x-a2)...(x-an)) .Thrn i took log on both sides and then differentiated both sides with respect to x.I got 1/y dy/dx=n/x -1/(x-a1)-1/(x-a2)...-1/(x-an).But now I was stuck.How will i get rid of that n there as it is not in the answer.
Help please.

 

[epenguin]

Have you checked your formula carefully? The scheme makes it look like in the second term you should have /(x-a₁₎, the third a₂x/...

 

[Mark44]

Problem Statement: if y=1+a1/(x-1) +a2x((x-a1)(x-a2)) +a3x^2/((x-a1)(x-a2)(x-a3))+...+anx^n-1/((x-a1)(x-a2)...(x-an)).
Find dy/dx.

Per @epenguin's correction and adding a missing /, and written in LaTeX, the above should probably be
$y=1+a_1/(x-a_1) +a_2x/((x-a_1)(x-a_2)) +a_3x^2/((x-a_1)(x-a_2)(x-a_3))+...+a_nx^{n-1}/((x-a_1)(x-a_2)...(x-a_n))$

Sorry i haven't used LaTeX this time as i will first have to learn it.

It's pretty simple for the equation above.
Subscripts: a_1
Exponents: x^2 or x^{n - 1}
The braces are optional if the subscript or exponent is just a single character.
Fractions: \frac {a_2x}{(x - a_1)(x - a_2)}
The LaTeX expressions can be surrounded by pairs of # characters (inline LaTeX) or pairs of $ characters (standalone LaTeX). There's a link to our tutorial at the bottom left of the input pane.

Relevant Equations: Logarithmic differentiation

I'm not sure how log differentiation is relevant. You can't simplify the log of a sum of terms.

I first solved the first two terms

What do you mean by "solved the first two terms"?

and then i solved the resulting term with the third term and so on.

... and "solved the resulting term"?

At last i was left with x^n/((x-a1)(x-a2)...(x-an)) .Thrn i took log on both sides and then differentiated both sides with respect to x.I got 1/y dy/dx=n/x -1/(x-a1)-1/(x-a2)...-1/(x-an).But now I was stuck.How will i get rid of that n there as it is not in the answer.
Help please.

 

[Physics lover]

Per @epenguin's correction and adding a missing /, and written in LaTeX, the above should probably be
$y=1+a_1/(x-a_1) +a_2x/((x-a_1)(x-a_2)) +a_3x^2/((x-a_1)(x-a_2)(x-a_3))+...+a_nx^{n-1}/((x-a_1)(x-a_2)...(x-a_n))$
It's pretty simple for the equation above.
Subscripts: a_1
Exponents: x^2 or x^{n - 1}
The braces are optional if the subscript or exponent is just a single character.
Fractions: \frac {a_2x}{(x - a_1)(x - a_2)}
The LaTeX expressions can be surrounded by pairs of # characters (inline LaTeX) or pairs of $ characters (standalone LaTeX). There's a link to our tutorial at the bottom left of the input pane.
I'm not sure how log differentiation is relevant. You can't simplify the log of a sum of terms.
What do you mean by "solved the first two terms"?
... and "solved the resulting term"?

Thanks for editing the post.
I meant that i first wrote 1+x/(x-a_1)=x/(x-a₁) and then x/(x-a₁1)+a₂x^2/((x-a₁1)(x-a₂)) as x^2/((x-a₁)(x-a₂)) and further.

 

[Mark44]

Thanks for editing the post.
I meant that i first wrote 1+x/(x-a1)=x/(x-a1) and then x/(x-a1)+a2x^2/((x-a1)(x-a2)) as x^2/((x-a1)(x-a2)) and further.

OK, so it sounds like you're doing an induction argument. If you can show (by induction) that the original expression can be written as $y = \frac {x^n}{(x - a_1)(x - a_2) \dots (x - a_n)}$, then you can take the log of both sides and differentiate.

 

[Physics lover]

OK, so it sounds like you're doing an induction argument. If you can show (by induction) that the original expression can be written as $y = \frac {x^n}{(x - a_1)(x - a_2) \dots (x - a_n)}$, then you can take the log of both sides and differentiate.

I have done that i got the smae thing at last then i took log

 

[Physics lover]

Bu

OK, so it sounds like you're doing an induction argument. If you can show (by induction) that the original expression can be written as $y = \frac {x^n}{(x - a_1)(x - a_2) \dots (x - a_n)}$, then you can take the log of both sides and differentiate.

t on taking log i am stuck with 'n'term how to get rid of it

 

[Mark44]

But on taking log i am stuck with 'n'term how to get rid of it

How do you know the n term is supposed to vanish? Are you basing this on a posted solution? Do you know for a fact that the posted solution is correct?

Something to try: Let $y = 1 + \frac {a_1}{x - a_1} + \frac{a_2x}{(x - a_1)(x - a_2)} = \frac{x^2}{(x - a_1)(x - a_2)}$
Take the derivative in the normal way, and compare this with what you get using log differentiation. You might be able to get some insight into the larger problem.

 

[Physics lover]

How do you know the n term is supposed to vanish? Are you basing this on a posted solution? Do you know for a fact that the posted solution is correct?

Something to try: Let $y = 1 + \frac {a_1}{x - a_1} + \frac{a_2x}{(x - a_1)(x - a_2)} = \frac{x^2}{(x - a_1)(x - a_2)}$
Take the derivative in the normal way, and compare this with what you get using log differentiation. You might be able to get some insight into the larger problem.

I was comparing it with the answer given in my book.The answer given is (y/x) (a₁/(a₁-x)+a₂/(a₂-x)...+a_n/(a_n-x))

 

[pasmith]

But on taking log i am stuck with 'n'term how to get rid of it

Does the series terminate, or is it in fact infinite?

 

[Physics lover]

Does the series terminate, or is it in fact infinite?

It terminates with the last term being xⁿ/((x-a₁)(x-a₂)...(x-a_n))

 

[Physics lover]

How do you know the n term is supposed to vanish? Are you basing this on a posted solution? Do you know for a fact that the posted solution is correct?

Something to try: Let $y = 1 + \frac {a_1}{x - a_1} + \frac{a_2x}{(x - a_1)(x - a_2)} = \frac{x^2}{(x - a_1)(x - a_2)}$
Take the derivative in the normal way, and compare this with what you get using log differentiation. You might be able to get some insight into the larger problem.

Sorry but i can't get what you said.i differentiated that and i got (2a₁a₂y)/x

 

[Mark44]

I was comparing it with the answer given in my book.The answer given is (y/x) (a₁/(a₁-x)+a₂/(a₂-x)...+a_n/(a_n-x))

If I start with $y = \frac{x^2}{(x - a_1)(x - a_2)}$, I can use log differentiation to get to $y' = \frac y x(\frac{a_1}{a_1 - x} + \frac {a_2}{a_2 - x})$, so the textbook answer seems plausible to me.

The 'n' that you're struggling with is essential in breaking things up to get to the form in your textbook.
Take things a step further than I did, and start with $y = \frac {x^3}{(x - a_1)(x - a_2)(x - a_3)}$ (i.e., with the first four terms) to see how things are working.

 

[Mark44]

Sorry but i can't get what you said.i differentiated that and i got (2a₁a₂y)/x

If you differentiate $y = \frac {x^2}{(x - a_1)(x - a_2)}$, an intermediate step is
$\frac {y'}y = \frac 2 x + \frac 1 {a_1 - x} + \frac 1 {a_2 - x}$. Notice that I switched the orders of the 2nd and 3rd denominators on the right.

Going from here to what I posted in #13 took me about 5 steps, but it's really just algebra manipulation.

 

[Physics lover]

If you differentiate $y = \frac {x^2}{(x - a_1)(x - a_2)}$, an intermediate step is
$\frac {y'}y = \frac 2 x + \frac 1 {a_1 - x} + \frac 1 {a_2 - x}$. Notice that I switched the orders of the 2nd and 3rd denominators on the right.

Going from here to what I posted in #13 took me about 5 steps, but it's really just algebra manipulation.

Ok it got it from this i can take 1/x common and then write 2 as (a1-x)/(a1-x)+(a2-x)/(a2-x)

 

[Physics lover]

Ok it got it from this i can take 1/x common and then write 2 as (a1-x)/(a1-x)+(a2-x)/(a2-x)

Thanks for the help.I got the solution of big one also.