Question

If ${x^3} + {y^3} - 3axy = 0$ then $\dfrac{{dy}}{{dx}}$ equals to
A. $\dfrac{{ay - {x^2}}}{{{y^2} - ax}}$
B. $\dfrac{{ay - {x^2}}}{{ay - {y^2}}}$
C. $\dfrac{{ay + {x^2}}}{{ax + {y^2}}}$
D. None of these

Answer 1

Given question has simple algebraic expression of highest degree 3.We have to find the derivative of the function. It is very easy to solve only the basics of the derivative. We need to use the product rule for the third term of the equation since there are two functions $x$ and $y$ there.

Complete step by step answer:
Given the function is,
${x^3} + {y^3} - 3axy = 0$
Now taking the derivative as,
$\dfrac{d}{{dx}}{x^3} + \dfrac{d}{{dx}}{y^3} - 3a\dfrac{d}{{dx}}\left( {xy} \right) = 0$
We know that, $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
$3{x^2} + 3{y^2}\dfrac{{dy}}{{dx}} - 3a\left( {y.1 + x.\dfrac{{dy}}{{dx}}} \right) = 0$
Dividing both sides by 3 we get,
${x^2} + {y^2}\dfrac{{dy}}{{dx}} - a\left( {y + x.\dfrac{{dy}}{{dx}}} \right) = 0$
This is the next step that is obtained. Now we need to separate the $\dfrac{{dy}}{{dx}}$ terms. so on opening the bracket,
${x^2} + {y^2}\dfrac{{dy}}{{dx}} - ay - ax.\dfrac{{dy}}{{dx}} = 0$
$\Rightarrow {x^2} - ay = ax.\dfrac{{dy}}{{dx}} - {y^2}\dfrac{{dy}}{{dx}}$
Taking $\dfrac{{dy}}{{dx}}$ common,
${x^2} - ay = \dfrac{{dy}}{{dx}}\left( {ax - {y^2}} \right)$
Transposing the terms,
$\dfrac{{dy}}{{dx}} = \dfrac{{{x^2} - ay}}{{ax - {y^2}}}$
On taking minus common from numerator and denominator,
$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{ay - {x^2}}}{{{y^2} - ax}}$

Thus option A is correct.

Note: Here note that the equation is containing x and y terms only. The third term involves 3a which is constant. So it is taken outside the derivative. When we find the $\dfrac{{dy}}{{dx}}$ we might get confused that the option so found is not available in the options so given. But after making some changes we can see that the option is available. All we need to note is we should check for all the possible changes or transpositions in the options.