text: # differentiating with multiple variables
  
  • Apr 7th 2012, 02:00 PM
  rabert1
  differentiating with multiple variables
  $h(z) = 3cos(z) + 3^x + cos^{-1}(2x)$
  
  $h'(z) = (3cos(z))' + (3^x)' + (cos^{-1}(2x))'$
  
  $h'(z) = -3sin(z) + (3^x)' + (cos^{-1}(2x))'$ //how do I finish?
  • Apr 7th 2012, 02:23 PM
  emakarov
  Re: differentiating with multiple variables
  If x and z are independent variables, then the function should be written as h(x,z), and the derivative with respect to z should be denoted by $\frac{\partial h}{\partial z}$. In this case, x is treated a constant, so $\frac{\partial}{\partial z}(3^x + cos^{-1}(2x))=0$.
  • Apr 7th 2012, 02:41 PM
  rabert1
  Re: differentiating with multiple variables
  Thanks/
  So variables are constants, unless they are taken with respect to the function?
  • Apr 7th 2012, 03:02 PM
  emakarov
  Re: differentiating with multiple variables
  Quote:
  
  Originally Posted by rabert1
  So variables are constants, unless they are taken with respect to the function?
  
  Sorry, this sentence does not make much sense.
  • Apr 7th 2012, 03:28 PM
  Plato
  Re: differentiating with multiple variables
  Quote:
  
  Originally Posted by rabert1
  So variables are constants, unless they are taken with respect to the function?
  
  Have you studied implicit differentiation?
  
  Like a circle: $x^2+y^2=r^2$. We want to find $y'=\frac{dy}{dx}$.
  
  Thus $2x+2yy'=0$ solve for $y'$.
  • Apr 7th 2012, 06:52 PM
  rabert1
  Re: differentiating with multiple variables
  
  How would one know when the variable is a constant or not?
  See you have from the equation:
  
  $(y^2)' = 2yy'$
  
  and
  
  $(r^2)' = 0$
  Why wouldn't $(r^2)' = 2rr'$
  • Apr 8th 2012, 03:00 AM
  Plato
  Re: differentiating with multiple variables
  Quote:
  
  Originally Posted by rabert1
  See you have from the equation: $(y^2)' = 2yy'$
  $(r^2)' = 0$
  Why wouldn't $(r^2)' = 2rr'$
  One is expected to know that a circle is in two variables, usually $x~\&~y$.