# Differentiation question

Printable View

• Oct 15th 2008, 08:25 PM
alberty02
Differentiation question
The equation of a curve is xy(x+y)=2(a)^3 where a is a non-zero constant. show that there is only one point on the curve at which the tangent is parallel to the x-axis, and find the coordinates of this point.

can some1 please help me with this question please.
thanks in advance
• Oct 15th 2008, 08:47 PM
o_O
$\begin{array}{ccl}xy(x+y) & = & 2a^3 \\ x^2y + xy^2 & = & 2a^{3} \\ {\color{red}(x^2y)'} + {(\color{blue} xy^2)' } & = & {\color{magenta}(2a^3)'} \\ {\color{red}\left[(x^2)'y + x^2(y)'\right]} + {\color{blue} \left[ (x)'y^2 + x(y^2)'\right]} & = & {\color{magenta}0} \end{array}$

So far, I've multiplied in the xy into the brackets. Then, I differentiate both sides. Since y is a function of x, we have to use product rule on each product on the left hand side. On the right hand side, since a is just a constant, its derivative is simply 0.

See where you can go from this and come back with any more questions.