We're given z equals x cubed plus xy squared,
where x equals uv cubed and y equals u plus v times e
raised to the power of w.
We're asked to determine the partial derivative of z
with respect to u when u equals two, v equals negative one,
and w equals zero.
Notice here the given function of z is a function
of two variables, where x and y are functions
of three variables, u, v and w.
And therefore to find this partial derivative,
we'll have to apply the chain rule,
which is given here below,
where if the given function z is a function
of two variables, which are functions of three independent
variables, we would use the chain rule to find
the partial derivatives given here below.
But notice in our notes, the independent variables
are s, t, and u, and in our problem,
the independent variables are u, v, and w.
So let's being by setting up the chain rule
for the partial derivative of z with respect to u.
We'd have the partial derivative of z with respect to u
is equal to the partial of z with respect to x,
times the partial of x with respect to u,
and then we'd have plus the partial of z with respect to y,
times the partial of y with respect to u.
So we'll first determine this partial derivative,
then evaluate at the given values for v and w.
So the partial of z with respect to u is equal to
the partial of z with respect to x,
so differentiate z with respect to x
treating y as a constant,
so we would have 3x squared plus y squared.
Again, we're treating y as a constant,
as we did for differentiate with respect to x.
And then we have times the partial of x with respect to u,
so here's x, we differentiate x with respect to u,
treating v as a constant,
and therefore we just have v cubed.
Then we have plus the partial of z with respect to y,
which we differentiate z with respect to y,
treating x as a constant, and therefore,
the derivative would just be 2xy.
And then finally we have times the partial of y,
with respect to u,
so differentiate y with respect to u,
we treat both v and w as constants,
and therefore the partial derivative is just one.
Now from here, if we were asked to give the partial of z
with respect to u, we would want this function in terms
of just u, v, and w, but for this question,
because we're just evaluating it,
at u, v, and w, instead of performing substitutions
for x and y, let's go ahead and leave it in this form,
and what we'll do is determine the value of x and y
when u equals 2, v equals negative one, and w equals zero.
That'll make the substitution a lot more manageable.
So let's find x of two comma negative one comma zero,
meaning the value of x when u equals two,
v equals negative one, and w equals zero.
So here's x, so we have u which is two,
times v cubed which would be negative one cubed,
which would just be negative two.
Now let's find y of two comma negative one comma zero.
So y is equal to u plus v times e to the w,
so we'd have two plus negative one,
times e raised to the power of zero,
e to the zero is one, so we just have two plus negative one,
which equals one.
So now we'll determine the value of the partial derivative.
Again, when u equals two, v equals negative one,
and w equals zero.
So we have three and then for x squared we know x
is going to be negative two, so we have negative two squared
plus y squared which would be one squared,
so we have this quantity times v cubed,
but we know v is negative one, so times negative one cubed,
and then plus, well 2xy times one is just 2xy,
so we have two times x which would be negative two,
times y which would be positive one.
So let's go ahead and simplify.
Working inside the parentheses here,
negative two squared is four,
four times three is 12,
12 plus one squared that'd be 12 plus one is 13,
and we have negative one cubed,
that's just going to be negative one,
and we have plus two times negative two times one
is negative four, so we have negative 13 plus negative four
which equals negative 17.
So we now know that the partial derivative of z
with respect to u is equal to negative 17
when u equals two, v equals negative one, and w equals zero.
I hope you found this helpful.