$\displaystyle h(z) = 3cos(z) + 3^x + cos^{-1}(2x)$
$\displaystyle h'(z) = (3cos(z))' + (3^x)' + (cos^{-1}(2x))' $
$\displaystyle h'(z) = -3sin(z) + (3^x)' + (cos^{-1}(2x))'$ //how do I finish?
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 $\displaystyle \frac{\partial}{\partial z}(3^x + cos^{-1}(2x))=0$.
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 $\displaystyle (r^2)' = 2rr'$