Sunday, February 28, 2016

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).

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