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

• Physics lover
In summary: 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 by doing this.
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.

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

Physics lover said:
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))##
Physics lover said:
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 $characters (standalone LaTeX). There's a link to our tutorial at the bottom left of the input pane. Physics lover said: Relevant Equations: Logarithmic differentiation I'm not sure how log differentiation is relevant. You can't simplify the log of a sum of terms. Physics lover said: I first solved the first two terms What do you mean by "solved the first two terms"? Physics lover said: and then i solved the resulting term with the third term and so on. ... and "solved the resulting term"? Physics lover said: At last i was left with x^n/((x-a1)(x-a2)...(x-an)) .Thrn i took log on both sides and then differentiated both sides with respect to x.I got 1/y dy/dx=n/x -1/(x-a1)-1/(x-a2)...-1/(x-an).But now I was stuck.How will i get rid of that n there as it is not in the answer. Help please. Mark44 said: Per @epenguin's correction and adding a missing /, and written in LaTeX, the above should probably be ##y=1+a_1/(x-a_1) +a_2x/((x-a_1)(x-a_2)) +a_3x^2/((x-a_1)(x-a_2)(x-a_3))+...+a_nx^{n-1}/((x-a_1)(x-a_2)...(x-a_n))## It's pretty simple for the equation above. Subscripts: a_1 Exponents: x^2 or x^{n - 1} The braces are optional if the subscript or exponent is just a single character. Fractions: \frac {a_2x}{(x - a_1)(x - a_2)} The LaTeX expressions can be surrounded by pairs of # characters (inline LaTeX) or pairs of$ 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.

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Physics lover said:
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.

Mark44 said:
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

Bu
Mark44 said:
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

Physics lover said:
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.

Mark44 said:
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)...+an/(an-x))

Physics lover said:
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?

pasmith said:
Does the series terminate, or is it in fact infinite?
It terminates with the last term being xn/((x-a1)(x-a2)...(x-an))

Mark44 said:
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

Physics lover said:
I was comparing it with the answer given in my book.The answer given is (y/x) (a1/(a1-x)+a2/(a2-x)...+an/(an-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.

Physics lover said:
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.

Mark44 said:
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 said:
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.

## 1. What is the purpose of finding the derivative of a series in calculus?

The derivative of a series helps us to find the rate of change of a function at a specific point. This is useful in many real-world applications, such as calculating velocity or acceleration in physics, or finding optimal solutions in economics and engineering.

## 2. How do you solve for x^n/((x-a1)(x-a2)...(x-an)) in calculus?

To solve for x^n/((x-a1)(x-a2)...(x-an)), we can use the quotient rule, which states that the derivative of a quotient is equal to (denominator * derivative of numerator - numerator * derivative of denominator) / (denominator)^2. We can also use the power rule and the chain rule to simplify the equation before applying the quotient rule.

## 3. Can the derivative of a series be negative?

Yes, the derivative of a series can be negative. This indicates that the function is decreasing at that point. The sign of the derivative is important in determining the behavior of a function, such as finding local maxima and minima.

## 4. Are there any special cases when finding the derivative of a series?

Yes, there are some special cases to consider when finding the derivative of a series. These include when the denominator is equal to zero, which would result in an undefined derivative, and when the exponent is negative, which would require the use of the power rule in reverse.

## 5. How is finding the derivative of a series related to the concept of limits?

Finding the derivative of a series is closely related to the concept of limits. In fact, the derivative is defined as the limit of the difference quotient as the change in x approaches zero. This means that the derivative is the instantaneous rate of change at a specific point, which is determined by the limiting behavior of the function as x approaches that point.

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