Question

If ${x^3} + {y^3} = 3axy$, find $\dfrac{{dy}}{{dx}}$

Answer 1

Hint: We will first differentiate the left -hand-side of the given equation using the formula $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ and differentiate the right -hand-side using the product formula ${\left( {f\left( x \right)g\left( x \right)} \right)^\prime } = f{\left( x \right)^\prime }g\left( x \right) + f\left( x \right)g{\left( x \right)^\prime }$. We will then rearrange the equation such that terms containing $\dfrac{{dy}}{{dx}}$ are on one side and hence find the value of $\dfrac{{dy}}{{dx}}$.

Complete step by step Answer:

Let given the equation ${x^3} + {y^3} = 3axy$.........(1)
Differentiate the given equation with respect to $x$.
The formula of differentiation of ${x^n}$ is $n{x^{n - 1}}$
We will use product rule in R.H.S of the equation.
The product rule states that ${\left( {f\left( x \right)g\left( x \right)} \right)^\prime } = f{\left( x \right)^\prime }g\left( x \right) + f\left( x \right)g{\left( x \right)^\prime }$
After applying the rules of derivative, we get,
$
  3{x^2} + 3{y^2}\dfrac{{dy}}{{dx}} = 3a\left( {x\dfrac{{dy}}{{dx}} + y} \right) \\
   \Rightarrow 3{x^2} + 3{y^2}\dfrac{{dy}}{{dx}} = 3ax\dfrac{{dy}}{{dx}} + 3ay \\
 $
Now, we will rearrange the equation and bring the terms of $\dfrac{{dy}}{{dx}}$ to one side.
$
  3{y^2}\dfrac{{dy}}{{dx}} - 3ax\dfrac{{dy}}{{dx}} = 3ay - 3{x^2} \\
   \Rightarrow \left( {3{y^2} - 3ax} \right)\dfrac{{dy}}{{dx}} = 3ay - 3{x^2} \\
 $
Taking 3 common and cancelling from both the sides
$
  3\left( {{y^2} - ax} \right)\dfrac{{dy}}{{dx}} = 3\left( {ay - {x^2}} \right) \\
   \Rightarrow \left( {{y^2} - ax} \right)\dfrac{{dy}}{{dx}} = \left( {ay - {x^2}} \right) \\
$
Now we can find the derivative by dividing \[\left( {{y^2} - ax} \right)\]
Hence,
\[\dfrac{{dy}}{{dx}} = \dfrac{{ay - {x^2}}}{{{y^2} - ax}}\]

Note: Students should remember the formulas of derivatives. Some students make mistakes in applying product formulas. The given equation is an implicit equation, that is we cannot express $y$ in terms of $x$.