differentiating with multiple variables

**rabert1** · April 7th 2012, 02:00 PM

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

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

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

**emakarov** · April 7th 2012, 02:23 PM

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 $\frac{\partial}{\partial z}(3^x + cos^{-1}(2x))=0$ .

**rabert1** · April 7th 2012, 02:41 PM

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

**emakarov** · April 7th 2012, 03:02 PM

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.

**Plato** · April 7th 2012, 03:28 PM

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 $y'$ .

**rabert1** · April 7th 2012, 06:52 PM

thanks for replying!

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 $(r^2)' = 2rr'$

**Plato** · April 8th 2012, 03:00 AM

Originally Posted by rabert1

thanks for replying!
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 $(r^2)' = 2rr'$

One is expected to know that a circle is in two variables, usually $x~\&~y$ .

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