Understanding Wolfram's Partial Derivative Widget

In summary, the speaker is seeking clarification on how Wolfram achieved the correct answer for a partial derivative calculation using the Wolframalpha widget. They are new to the site and do not plan to use any code. They provide a link to the widget and ask for help with a specific function. They also mention using Latex to write up their work. However, they are having trouble with a subsection of the problem involving factoring out constants and differentiating. They acknowledge that it may seem like an elementary question.
  • #1
Doctorchaos
1
0
Hello,

Wolfram is giving me the required answer however, the steps it uses I find very confusing. Can anyone share some light on how wolfram achieved the correct answer.

As I am new to this site, I won't be using any code. I am in the process of writing it up on Latex.

Here is the link for wolframs partial derivative widget

http://www.wolframalpha.com/widget/...tep solution&showAssumptions=1&showWarnings=1

However it keeps changing my function, if the exponential appears on the bottom, can you please use this function

(((1)\(y)^(0.5))))(e^((-x^2-4y^2)\(4y))) WITH RESPECT TO Y


Thankyou everyone, it might seem like an elementary question. The bit I don't understand though is where it says factor out the constants. Its not the factor out the constants but, but the differentiation inside that subsection. That's my main problem
 
Physics news on Phys.org
  • #2
Doctorchaos said:
Hello,

Wolfram is giving me the required answer however, the steps it uses I find very confusing. Can anyone share some light on how wolfram achieved the correct answer.

As I am new to this site, I won't be using any code. I am in the process of writing it up on Latex.

Here is the link for wolframs partial derivative widget

http://www.wolframalpha.com/widget/...tep solution&showAssumptions=1&showWarnings=1

However it keeps changing my function, if the exponential appears on the bottom, can you please use this function

(((1)\(y)^(0.5))))(e^((-x^2-4y^2)\(4y))) WITH RESPECT TO Y
What does the expression above have to do with the one you entered into Wolframalpha?
The function you entered into WA was
$$ \frac{\sqrt{x}}{e^{\sqrt{x}(y^2 + 4x^2)}}$$

Which one are you asking about?

If it's this one -- (((1)\(y)^(0.5))))(e^((-x^2-4y^2)\(4y))) -- the problem we most often run into is the OP using too few parentheses. Here you are using way too many, which makes what you wrote difficult to read.

In a simpler form, it would be (1/y.5)e((-x2 - 4y2)/(4y)

LaTeX can be used for complicated expressions, as below:
$$ \frac{1}{y^{1/2}}e^{\frac{-x^2 - 4y^2}{4y}}$$


Doctorchaos said:
Thankyou everyone, it might seem like an elementary question. The bit I don't understand though is where it says factor out the constants. Its not the factor out the constants but, but the differentiation inside that subsection. That's my main problem
 
Last edited:

1. What is a partial derivative?

A partial derivative is a mathematical concept used in calculus to measure the rate of change of a function with respect to one of its variables while holding all other variables constant. It is denoted by ∂ (pronounced "partial") and is often used in multivariable calculus to analyze functions with multiple independent variables.

2. How is a partial derivative calculated?

A partial derivative is calculated by taking the derivative of a function with respect to one variable while treating all other variables as constants. This can be done by following the same rules of differentiation as in single variable calculus, but treating the other variables as constants. For example, the partial derivative of f(x,y) with respect to x would be written as ∂f/∂x and calculated by differentiating with respect to x while treating y as a constant.

3. What is the difference between a partial derivative and a total derivative?

A partial derivative measures the rate of change of a function with respect to one variable while holding all other variables constant. A total derivative, on the other hand, measures the overall rate of change of a function with respect to all its variables. In other words, a total derivative takes into account the effects of all variables on the function, while a partial derivative only considers the effect of one variable at a time.

4. Can a partial derivative be negative?

Yes, a partial derivative can be negative. This means that the function is decreasing in value with respect to that variable. However, it is important to note that a negative partial derivative does not necessarily mean that the overall function is decreasing, as it only measures the rate of change with respect to one variable while holding all others constant.

5. What are the applications of partial derivatives?

Partial derivatives have various applications in fields such as physics, economics, and engineering. They are used to analyze how a function changes in response to changes in its variables, which can be useful in predicting and understanding real-world phenomena. For example, partial derivatives are used in economics to analyze the impact of different factors on a market, and in physics to calculate rates of change in complex systems.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
978
Back
Top