We want to find the first order of partial derivatives
of the given function F of X comma Y.
Let's first write the function F of X comma Y,
using a rational exponent where we can write
the square root as an exponent of one half.
So the given function is F of X comma Y
equals four times the quantity of five X cubed,
plus four Y, plus eight X Y to the fifth,
raised to the power of positive one half.
In order to find the partial derivatives,
we will have to apply the chain rule where
the inner function is five X cubed plus four Y
plus eight X Y the fifth, which we will call U.
So now we can think of the given function as just
four U to the one and half where again
U is equal to the inner function of five X cubed
plus four Y plus eight X Y to the fifth.
To find the partial of F with respect to X,
we consider Y a constant
and differentiate with respect to X.
So again, thinking of the function
as four U to the one half,
the derivative with respect to X is going to be four,
times one half, times U to the power of one half minus one
which is an exponent of negative one half, times U prime.
So U to the power of negative one half
is the quantity of five X cubed
plus four Y, plus eight X Y to the fifth,
raised to the power of negative one half,
and then to find U prime, we differentiate five X cubed
plus four Y plus eight X Y to the fifth,
with respect to X treating Y as a constant.
So we have, times the derivative of five X cubed
with respect to X is 15 X squared
because we're treating Y as a constant,
the derivative of 4 Y with respect to X is zero.
And then we have plus a derivative of
eight X Y to the fifth
with respect to X treating Y to the fifth is a constant.
And since the derivative of eight X with respect to X
is eight, we have plus eight Y to the fifth.
And now let's multiply
and write the partial derivative using positive exponents.
Remember if we move the quantity five X cubed, plus four Y
plus eight X Y to the fifth, raise the power of it
to negative one half down to the denominator,
it will change the exponent to positive one half.
So in the numerator we have four times one half
which is two, times the quantity 15 X squared
plus eight Y to the fifth.
In the denominator, we're going to have this
quantity raised to the power of one half,
which is equivalent to the square root.
So let's write the denominator as the square root
of the quantity five X cubed, plus four Y plus
eight X Y to the fifth.
And now to find the partial of F with respect to Y
we consider X constant and differentiate with respect to Y.
So again, thinking of the function in terms of U,
when you differentiate four U to the power of one half,
with respect to Y, treating X as a constant.
So we're still going to have four times one half,
times U to the power of one half minus one,
which gives us the quantity five X cubed
plus four Y, plus eight X Y to the fifth
raised the power of negative one half,
and then times U prime
where now U prime is derivative of five X cubed
plus four Y plus eight X Y to the fifth
with respect to Y treating X as a constant.
And therefore the derivative of a five X cubed with respect
to Y is equal to zero because we're treating X as a constant
plus the derivative of four Y with respect to Y is four.
So we have times of quantity, four plus the derivative of
eight X Y to the fifth
with respect to Y treating X as a constant.
So we multiply by five
which gives us 40 giving us 40 X Y to the power of four.
So again, let's go ahead and multiply
and write the partial derivative using a positive exponent
in the numerator we have four times one half
which is two, times the quantity four
plus 40 X Y to the fourth.
And again, moving the negative exponent
down to the denominator.
The exponent becomes positive one half
which is equivalent to the square root.
So we have the square root of the quantity of five X cubed
plus four Y plus eight X Y to the fifth in the denominator.
I hope you found this helpful.