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

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

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

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- Apr 7th 2012, 02:00 PMrabert1differentiating 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? - Apr 7th 2012, 02:23 PMemakarovRe: 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 $\displaystyle \frac{\partial}{\partial z}(3^x + cos^{-1}(2x))=0$.

- Apr 7th 2012, 02:41 PMrabert1Re: differentiating with multiple variables
Thanks/

So variables are constants, unless they are taken with respect to the function? - Apr 7th 2012, 03:02 PMemakarovRe: differentiating with multiple variables
- Apr 7th 2012, 03:28 PMPlatoRe: differentiating with multiple variables
- Apr 7th 2012, 06:52 PMrabert1Re: differentiating with multiple variables
thanks for replying!

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 $\displaystyle (r^2)' = 2rr'$ - Apr 8th 2012, 03:00 AMPlatoRe: differentiating with multiple variables