Partial derivative

Question

find the ∂f/∂x and ∂f/∂y of:

1.f(x,y)=tan^-1(y/x)

2.f(x,y)=e^-xsin(x+y)

3. f(x,y)=x/(x^2+y^2)

 

Answer 1

1.  The derivative of tan-1 (z) = 1/z2 * z' (the z' comes from the chain rule)

 

df/dx = 1/ (y/x)² * d/dx (y * (1/x))

 

df/dx = 1/ (y²/x²) * (-2y/x²)

 

df/dx = (x²/y²) * (-2y/x²)

 

df/dx = x²*-2y

             x²*y²

 

df/dx =  - 2

              y

 

df/dy = 1/(y/x)² * d/dx(y/x)

 

df/dy = 1/(y²/x²) * (1/x)

 

df/dy = (x²/y²) * (1/x)

 

df/dy =     x²  

              y² x

 

df/dy =    x 

              y²

 

2.  df/dx = e^-xcos(x+y) + sin(x+y)(-e^-x)  (use the product rule)

 

df/dx = e^-xcos(x+y) - e^-xsin(x+y)

 

df/dy = e^-xcos(x+y)

 

You don't need the product rule to get df/dy because e^-x is a "constant" with respect to y.  It's just a number, rather than a function.  If it was e^y, then you would need to use the product rule.

 

3.  Use the quotient rule to get d/dx.

 

df/dx = (x²+y²)(1) - (x)(2x)

                  (x²+y²)²

 

 df/dx =    y² - x²
              (x²+y²)²

 

To get d/dy, you treat the x and x² terms as constants.  You still need the quotient rule because the y² is in the denominator:

 

d/dy = (x²+y²)(0) - (x*2y)

                (x²+y²)²

 

d/dy =       2xy    

             (x²+y²)²