What is value of the derivative of x^3 + y^3 = 14xy at (7,7)?

Given the equation:

x^3 + y^3 =
14xy

We need to find the value of the first derivative at the point
(7,7).

First we will use implicit
differentiation.

==> (x^3)' + (y^3)' =
14(xy)'

==> 3x^2 + 3y^2 *y' = 14[
(x'y+xy')]

==> 3x^2 + 3y^2 y' = 14(y +
xy')

==> 3x^2 + 3y^2 y' = 14y +
14xy'

Now we will combine terms with
y'.

==> 3y^2 y' - 14xy' = 14y -
3x^2

==> y' ( 3y^2-14x) = 14y -
3x^2

==> y' = (14y-3x^2)/
(3y^2-14x)

Now we will substitute with
(7,7)

==> y'(7,7) = (14*7 - 3*7^2) / (3*7^2 -
14*7)

= 98-147 / 147-98

= -49/49 =
-1

Then the value of the derivative is
(-1).