# 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 $x~\&~y$.