Partial differentiation find df/dx and df/dy

In summary, to find the derivatives of the function f(x,y)=tan^-1(y/x), you would use the chain rule and the derivative of tan^-1(u) is 1/(1+u^2). Using this, the derivatives with respect to x and y would be -y/x^2 and 1/(1+y^2/x^2) respectively.
  • #1
fionamb83
8
0

Homework Statement


For the function of two variables f(x,y)=tan^-1(y/x)
find df/dx and df/dy

I know i just differentiate with respect to x and then to y but I'm stuck on the tan^-1(y/x)
I know tan^-1(x)=1/1+X^2 when I applied this with respect to x I get 1/-1+y
I think this is wrong please help!
 
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  • #2
You should use the chain rule

[tex]
\frac{\partial}{\partial x}\tan^{-1}\left(\frac{y}{x}\right)=\left.\frac{d}{du}\tan^{-1}(u)\right|_{u=\frac{y}{x}}\frac{\partial}{\partial x}\frac{y}{x}
[/tex]
 
  • #3
fionamb83 said:
I know tan^-1(x)=1/1+X^2 ...
That's not true.
What is true is that
[tex]\frac{d}{dx} tan^{-1}(x) = \frac{1}{1 + x^2}[/tex]
 
  • #4
The derivative of tan-1(u) is
[itex]\frac{1}{1+ u^2}[/itex]
with u= y/x, that is
[itex]\frac{1}{1+ \frac{y^2}{x^2}}[/itex]
for the derivative with respect to x or y those are multiplied by the derivative of y/x with respect to x and the derivative of y/x with respect to x "respectively".
 
  • #5
so df/dx = -2x/1+y^2 ?
 
  • #6
fionamb83 said:
so df/dx = -2x/1+y^2 ?

How did you get that? Halls told you it's 1/(1+y^2/x^2) times the x derivative of y/x. What's the x derivative of y/x?
 
  • #7
Sorry just saw what I did there. Oops. the x derivative of y/x is -y/x^2. Sorry confused myself there. Thanks for the help everyone!
 

FAQ: Partial differentiation find df/dx and df/dy

What is partial differentiation?

Partial differentiation is a mathematical concept used in multivariable calculus. It involves finding the rate of change of a function with respect to one of its variables while holding all other variables constant.

How do you find the partial derivative df/dx?

To find the partial derivative df/dx, you first need to determine the function's total derivative with respect to x. Then, you can simply differentiate the function with respect to x while treating all other variables as constants.

How is partial differentiation useful in science?

Partial differentiation is commonly used in physics, engineering, and other scientific fields to analyze systems with multiple variables. It allows us to understand how the change in one variable affects the overall behavior of a system.

Can you give an example of a real-world application of partial differentiation?

One example of a real-world application of partial differentiation is in economics, where it is used to analyze the relationship between two or more variables, such as the demand and supply of a product.

Is there a difference between partial differentiation and normal differentiation?

Yes, there is a difference between partial differentiation and normal differentiation. Partial differentiation involves finding the rate of change of a function with respect to one variable while holding all other variables constant, whereas normal differentiation involves finding the rate of change of a function with respect to one variable without any constraints on the other variables.

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