We want to find the maximum rate of change
of the function f of x comma y equals four y
to the fifth divided by x at the point four comma one.
So the first thing to remember here is the gradient
of f is going to give us the direction
of maximum rate of change of f at the point four comma one,
but defining the actual maximum rate of change,
we'll need to determine the magnitude of the gradient
of f at the point four comma one.
So again for a quick review,
the direction of maximum increase of f is given by
the gradient of f at the point x comma y,
while the maximum value of the directional derivative
of f at the point x comma y,
or the maximum rate of increase,
is equal to the magnitude of the gradient of f
at the point x comma y.
So we'll have two steps here.
We'll first find the gradient at the point four comma one,
and then find the magnitude of that two-dimensional vector.
And again the given function is f of x comma y
equals four y to the fifth divided by x,
but if we wanted to, we could move this x
up to the numerator which would change the sign
of the exponent.
We could write this as four x to the negative one
y to the fifth.
And now we'll find the gradient of f,
also referred to as del f.
So we'll find partial of f with respect to x,
and the partial of f with respect to y.
So to find the partial of f with respect to x,
we're going to differentiate four x to the negative one
y to the fifth with respect to x treating y as a constant,
so the partial derivative would be,
we'd multiply by negative one,
so negative four, x to the power of,
subtract one from the exponent,
so that's negative two y to the fifth.
And now to find the partial with respect to y,
we differentiate with respect to y,
treating x as a constant.
So we'd multiply by five.
That'd be 20 x to negative one,
subtract one from the exponent of y,
which gives us y to the fourth.
And of course we could also express this as
an x component of negative four y to the fifth
over x to the second,
and we express the y component as 20 y to the fourth over x.
Either way, we now need to evaluate this
at the point four comma one.
So now we'll find the gradient of f, or del f,
at the point four comma one.
Let's go ahead and use this form here.
So now we'll substitute
four for x and one for y.
So the x component would be negative four
times one to the fifth
divided by four squared,
and the y component would be 20 times one to the fourth
divided by x which is four.
So simplifying this would be negative four over 16,
or negative one-fourth,
and here we just have 20 divided by four which equals five.
And now the maximum rate of change
at the point four comma one is going to be equal
to the magnitude
of the gradient,
so now we need to find the magnitude of our gradient
at the point four comma one.
So this is equal to the square root
of, we'd have negative one-fourth squared plus five squared.
This would be the square root of one-sixteenth
plus 25 or 25 over one if we want to find the exact value.
The common denominator would be 16 here,
so this is going to be equal to the square root of,
our denominator is going to be 16,
and for the numerator,
25 times 16 is equal to 400.
400 plus one is 401.
And since the square root of 16 is equal to four,
the exact value of the magnitude
would be equal to the square root of 401 divided by four.
Let's also get a decimal approximation.
The square root of 401,
right arrow,
divided by four.
Now if we did not need the exact value, though,
notice how we could just enter the square root
of one-sixteenth,
so one divided by 16
plus 25,
and we get the same decimal approximation.
So to four decimal places, this would be approximately
five point zero zero six two,
which again is the maximum rate of change of f
at the point four comma one.
Let's also look at this graphically.
So here's the graph of our surface.
We were given the point four comma one,
which is this point here in the xy plane.
The second red point here would be the corresponding
point on the surface.
To find the z coordinate of this point,
we have to substitute the x and y values
from this point into the function,
and from here we found the gradient
at the point four comma one,
which is this two-dimensional vector here in the xy plane.
This vector points in the direction of the maximum rate
of increase on the surface from this point,
but our goal was to find the maximum rate of change,
which is the magnitude of this vector here.
So the magnitude of this vector here
is the value that we found
as approximately five point zero zero six two.
Notice how I've also graphed the tangent line to the surface
in the same direction as the gradient,
which means the value that we found,
the magnitude of the gradient,
is also the slope of this tangent line to the surface.
Again, in the same direction as the gradient.
I hope you found this helpful.