# Thread: differentiating with multiple variables

1. ## differentiating with multiple variables

$\displaystyle h(z) = 3cos(z) + 3^x + cos^{-1}(2x)$

$\displaystyle h'(z) = (3cos(z))' + (3^x)' + (cos^{-1}(2x))'$

$\displaystyle h'(z) = -3sin(z) + (3^x)' + (cos^{-1}(2x))'$ //how do I finish?

2. ## 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 $\displaystyle \frac{\partial h}{\partial z}$. In this case, x is treated a constant, so $\displaystyle \frac{\partial}{\partial z}(3^x + cos^{-1}(2x))=0$.

3. ## Re: differentiating with multiple variables

Thanks/
So variables are constants, unless they are taken with respect to the function?

4. ## Re: differentiating with multiple variables

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.

5. ## Re: differentiating with multiple variables

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: $\displaystyle x^2+y^2=r^2$. We want to find $\displaystyle y'=\frac{dy}{dx}$.

Thus $\displaystyle 2x+2yy'=0$ solve for $\displaystyle y'$.

6. ## Re: differentiating with multiple variables

How would one know when the variable is a constant or not?
See you have from the equation:

$\displaystyle (y^2)' = 2yy'$

and

$\displaystyle (r^2)' = 0$
Why wouldn't $\displaystyle (r^2)' = 2rr'$

7. ## Re: differentiating with multiple variables

Originally Posted by rabert1
See you have from the equation:$\displaystyle (y^2)' = 2yy'$
$\displaystyle (r^2)' = 0$
Why wouldn't $\displaystyle (r^2)' = 2rr'$
One is expected to know that a circle is in two variables, usually $\displaystyle x~\&~y$.