# How to solve using Leibniz Rule

• Md. Abde Mannaf
In summary, the Leibniz rule in one dimension would be:$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x)) b'(x) - f(a(x)) a'(x)$$
Md. Abde Mannaf

## The Attempt at a Solution

here is i am still stuck.

berkeman
I believe there are quite a few typos in the problem statement if you are intending to use Leibniz rule. I for one have never seen an ##\text{In}## function before.

The Leibniz rule in one dimension would be:

$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x)) b'(x) - f(a(x)) a'(x)$$

EDIT: It would be more appropriate to call this the fundamental theorem.

Last edited:
Md. Abde Mannaf
at first i am use substitution. some one said
Don't use substitution, use Leibniz rule with α as the second variable.

what am i do now?

i know the Leibniz's rule. i solve many mathematical term . bt i cann't solve this math with Leibniz rule. i am still stuck
here. here limit is constant so 2nd and 3rd term will be zero.

any one solve this mathematical term and prove it??

Md. Abde Mannaf said:
i know the Leibniz's rule. i solve many mathematical term . bt i cann't solve this math with Leibniz rule. i am still stuck
here. here limit is constant so 2nd and 3rd term will be zero.

Your application of Leibnitz' rule is correct. You could try to change variables to ##u = \cos(x)## in your integral.

Md. Abde Mannaf
Zondrina said:
if you are intending to use Leibniz rule. I for one have never seen an In\text{In} function before.
See this, example 3.
Now let's do that daunting looking integral. Like I said before, this is a good problem.
First write ##\int_0^{\pi} \frac{1}{\alpha}-\frac{1}{(\alpha)(1+\alpha cos(x))} dx## for the original integral.
However ##\int \frac{1}{1+\alpha cos(x)} dx## itself is not nice, in-fact it's pretty nasty. It will be an interesting exercise to do, but even after that there is a good bit to do for the original problem. So I would recommend you use Wolfram for that for now, and proceed.

Md. Abde Mannaf
i could not solve this math above analysis . i am trying to my best. but i am fail every time. and again try...

please see this and give me more idea to solve

Md. Abde Mannaf said:

## The Attempt at a Solution

here is i am still stuck.
You can even cheat :) using wolframalpha.com.
The substitution tan(x/2)=u works well if you have a rational expression of trigonometric functions. Try.

Md. Abde Mannaf
Zondrina said:
I believe there are quite a few typos in the problem statement if you are intending to use Leibniz rule. I for one have never seen an ##\text{In}## function before.

The Leibniz rule in one dimension would be:

$$\frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt = f(b(x)) b'(x) - f(a(x)) a'(x)$$

EDIT: It would be more appropriate to call this the fundamental theorem.

The derivation of the OP is correct. Read Leibnitz Rule http://en.wikipedia.org/wiki/Leibniz_integral_rule

Md. Abde Mannaf
In regards to post #9:- I'm not entirely sure what you're trying to do here. Did you have a look at the links I provided in post #8. I think your trying to come up with a value for ##\int_0^{\pi} \frac{1}{1+\alpha cos(x)} dx## for ##|\alpha|<1##using ##\int_0^{\pi} \frac{1}{\alpha-cos(x)} dx=\frac{\pi}{\sqrt{\alpha^2-1}}## for ##|\alpha|>1##, however that + sign in the original integral makes a good difference.
ehild said:
You can even cheat :) using wolframalpha.com.
Does no one read my posts :-)

Md. Abde Mannaf
Md. Abde Mannaf said:
i could not solve this math above analysis . i am trying to my best. but i am fail every time. and again try...

please see this and give me more idea to solve

You can write your integral for ##f'(\alpha)## as ##\int_0^{\pi/2} + \int_{\pi/2}^{\pi}##, then change variables to ##x \leftarrow \pi -x## in the second integral, to get
$$f'(\alpha) = \int_0^{\pi/2} \left[ \frac{\cos(x)}{1 + \alpha \cos(x)} - \frac{\cos(x)}{1 - \alpha \cos(x)} \right] \, dx \\ = -\int_0^{\pi/2} \frac{2 \alpha \cos^2(x)}{1 - \alpha^2 \cos^2(x)} \, dx$$
Changing variables to ##\tan(x) = y## produces an integral in ##y## that can be split up into partial fractions, giving two well-known and easily do-able integrations.

Md. Abde Mannaf
certainly said:
Now let's do that daunting looking integral. Like I said before, this is a good problem.
First write ##\int_0^{\pi} \frac{1}{\alpha}-\frac{1}{(\alpha)(1+\alpha cos(x))} dx## for the original integral.
However ##\int \frac{1}{1+\alpha cos(x)} dx## itself is not nice, in-fact it's pretty nasty.
##\int \frac{1}{1+\alpha cos(x)} dx## is not that nasty.
Use the identity ##\cos(x)=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}##. Substitute u=tan(x/2), x=2arctan(u), ##dx=\frac{2}{1+u^2}du##. The integration limits become 0-->infinite.

Md. Abde Mannaf and SammyS
Md. Abde Mannaf
ehild said:
##\int \frac{1}{1+\alpha cos(x)} dx## is not that nasty.
Use the identity ##\cos(x)=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}##. Substitute u=tan(x/2), x=2arctan(u), ##dx=\frac{2}{1+u^2}du##. The integration limits become 0-->infinite.
Yeah, your right. I didn't actually do the integral, I only saw the answer and thought, this might take some time.

Md. Abde Mannaf

## What is Leibniz Rule?

Leibniz Rule, also known as the generalized product rule, is a mathematical formula used to differentiate a product of two or more functions. It can be used to find the derivative of complicated functions that cannot be easily differentiated using basic differentiation rules.

## How do you use Leibniz Rule to solve?

To use Leibniz Rule, follow these steps:

1. Identify the functions that are being multiplied together.
2. Write out the formula for Leibniz Rule: d/dx(f(x) * g(x)) = f(x) * g'(x) + f'(x) * g(x)
3. Find the derivatives of each function, f'(x) and g'(x).
4. Plug these values into the formula.
5. Simplify the equation if possible.

## When should Leibniz Rule be used?

Leibniz Rule should be used when trying to find the derivative of a product of two or more functions. It is especially useful when working with complicated functions that cannot be easily differentiated using basic rules.

## What are the limitations of Leibniz Rule?

Leibniz Rule cannot be used to differentiate a product of an infinite number of functions. It also cannot be used if the functions are not differentiable or if they are not defined for all values of x.

## Can Leibniz Rule be used to find higher order derivatives?

Yes, Leibniz Rule can be extended to find higher order derivatives for products of multiple functions. The formula for finding the nth derivative is: d^n/dx^n(f(x) * g(x)) = f(x) * g^(n)(x) + f^(n)(x) * g(x) + sum(k=1 to n-1) [f^(k)(x) * g^(n-k)(x)].

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