So, there are some examples of partial derivatives. Hopefully you will agree
that as long as we can remember to treat the other variables as constants
these work in exactly the same manner that derivatives of functions of one
variable do. So, if you can do Calculus I derivatives you shouldn’t have too
much difficulty in doing basic partial derivatives.
There is one final topic that we need to take a quick look at in this section,
implicit differentiation. Before getting into implicit differentiation for multiple
variable functions let’s first remember how implicit differentiation works for
functions of one variable.
Example 3 Find dydxdydx for 3y4+x7=5x3y4+x7=5x.
Show Solution
Now, we did this problem because implicit differentiation works in exactly
the same manner with functions of multiple variables. If we have a function
in terms of three variables xx, yy, and zz we will assume that zz is in fact a
function of xx and yy. In other words, z=z(x,y)z=z(x,y). Then whenever we
differentiate zz’s with respect to xx we will use the chain rule and add on
a ∂z∂x∂z∂x. Likewise, whenever we differentiate zz’s with respect to yy we
will add on a ∂z∂y∂z∂y.
Let’s take a quick look at a couple of implicit differentiation problems.
Example 4 Find ∂z∂x∂z∂x and ∂z∂y∂z∂y for each of the following functions.
1. x3z2−5xy5z=x2+y3x3z2−5xy5z=x2+y3
2. x2sin(2y−5z)=1+ycos(6zx)x2sin⁡(2y−5z)=1+ycos⁡(6zx)
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a x3z2−5xy5z=x2+y3x3z2−5xy5z=x2+y3 Show Solution
b x2sin(2y−5z)=1+ycos(6zx)x2sin⁡(2y−5z)=1+ycos⁡(6zx) Show Solution