# I did not understand this derivative -- help please

• cemtu
In summary: Can someone else explain it to me?This...this is really confusing. Can someone else explain it to me?
cemtu
Homework Statement
Mathematical Methods in Physics
Relevant Equations
Derivative
I have no idea how this derivative was taken.

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It's wrong. ##N_{0}## is a constant (assuming the conventional interpretation of ##N_0## as the value of ##N## at ##t=0##). ##\frac{dN_0}{dt}## is always going to be ##0## and doesn't really make any sense. And to go one further, you still need to derive ##e^{-\lambda t}## wrt ##t##.

dN0/dt = 2 is given thus N0 is 2t. There is no problem there sir.

cemtu said:
dN0/dt = 2 is given thus N0 is 2t. There is no problem there sir.
Your original post makes no sense. If you want any help you'll have to organise it into something meaningful.

etotheipi
Okay right away.

PeroK said:
Your original post makes no sense. If you want any help you'll have to organise it into something meaningful.
I put the whole question and its solution.

Are you sure you don't mean ##A_{0} = \left[-\frac{dN}{dt} \right]_{t=0} = 2##?

Of course, start off with the equation you wrote of ##N = N_{0}e^{-\lambda t}##. Then you can derive both sides with respect to ##t## to obtain a relation pertaining to the activities.

How would you derive the RHS wrt ##t##? Or perhaps more specifically, what can we do about the constant out the front?

etotheipi said:
Are you sure you don't mean ##A_{0} = \left[-\frac{dN}{dt} \right]_{t=0} = 2##?

Of course, start off with the equation you wrote of ##N = N_{0}e^{-\lambda t}##. Then you can derive both sides with respect to ##t## to obtain a relation pertaining to the activities.

How would you derive the RHS wrt ##t##? Or perhaps more specifically, what can we do about the constant out the front?
I know how to derive however our professor just replaced N with dN/dt (LHS) and N0 with dN0/dt (RHS). I don't know how. He didnt took the derivative as the way you and I want.

cemtu said:
I know how to derive however our professor just replaced N with dN/dt (LHS) and N0 with dN0/dt (RHS). I don't know how. He didnt took the derivative as the way you and I want.

I can't speak for what your professor was trying to imply however I can say that ##\frac{dN_{0}}{dt}## is zero, as is the case for all constants. It's essentially a meaningless statement.

What do you get if you derive both sides of ##N = N_0 e^{-\lambda t}## with respect to ##t##?

etotheipi said:
I can't speak for what your professor was trying to imply however I can say that ##\frac{dN_{0}}{dt}## is zero, as is the case for all constants. It's essentially a meaningless statement.

What do you get if you derive both sides of ##N = N_0 e^{-\lambda t}## with respect to ##t##?
Here

Last edited by a moderator:
cemtu said:
Here
I suspect this might be an initial value problem. But, you've misunderstood what's been given as initial values.

Without seeing the problem stated clearly, we are all just guessing.

etotheipi
cemtu said:
Here

You've done the differentiation right, ##\frac{dN}{dt} = -\lambda N_{0} e^{-\lambda t} \implies A = A_{0}e^{-\lambda t}## since ##A = \lambda N##.

You must stop using ##\frac{dN_{0}}{dt}##. It is wrong. I strongly suspect the intended meaning is the rate of change of ##N## at ##t=0##, however we might just denote this (negative) ##A_{0}## (or that more clunky expression in post #7).

You should be able to just plug in the ratio of the activities.

cemtu
etotheipi said:
You've done the differentiation right, ##\frac{dN}{dt} = -\lambda N_{0} e^{-\lambda t} \implies A = A_{0}e^{-\lambda t}## since ##A = \lambda N##.

You must stop using ##\frac{dN_{0}}{dt}##. It is wrong. I strongly suspect the intended meaning is the rate of change of ##N## at ##t=0##, however we might just denote this (negative) ##A_{0}## (or that more clunky expression in post #7).

You should be able to just plug in the ratio of the activities.
Sir, thank you.

cemtu said:
Sir, thank you.

You're welcome, though no need to call me 'sir', I haven't met the Queen yet!

PeroK said:
Without seeing the problem stated clearly, we are all just guessing.
This...

## What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It is essentially the slope of a tangent line to the function at that point.

## Why is understanding derivatives important?

Derivatives are used in many areas of science and engineering to model and analyze various processes and phenomena. They are particularly useful in understanding how variables change over time and in optimizing functions.

## How do I calculate a derivative?

The most common method for calculating a derivative is using the power rule, which states that the derivative of a function is equal to the exponent multiplied by the coefficient, with the exponent decreased by 1. There are also other rules and techniques, such as the chain rule and product rule, for calculating derivatives of more complex functions.

## What are some applications of derivatives?

Derivatives have many applications in science and engineering, such as in physics (to calculate velocity and acceleration), economics (to determine marginal cost and revenue), and biology (to model population growth). They are also used in machine learning and artificial intelligence to optimize algorithms and make predictions.

## What are some common mistakes when working with derivatives?

Some common mistakes when working with derivatives include forgetting to apply the chain rule or product rule, making algebraic errors, and not simplifying the final answer. It is important to carefully follow the rules and double-check calculations to avoid these mistakes.

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