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Question: If \({x^3} + {y^3} = 3axy\), find \(\dfrac{{dy}}{{dx}}\)...

If x3+y3=3axy{x^3} + {y^3} = 3axy, find dydx\dfrac{{dy}}{{dx}}

Explanation

Solution

We will first differentiate the left -hand-side of the given equation using the formula ddx(xn)=nxn−1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} and differentiate the right -hand-side using the product formula (f(x)g(x))′=f(x)′g(x)+f(x)g(x)′{\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 dydx\dfrac{{dy}}{{dx}} are on one side and hence find the value of dydx\dfrac{{dy}}{{dx}}.

Complete step by step Answer:

Let given the equation x3+y3=3axy{x^3} + {y^3} = 3axy.........(1)
Differentiate the given equation with respect to xx.
The formula of differentiation of xn{x^n} is nxn−1n{x^{n - 1}}
We will use product rule in R.H.S of the equation.
The product rule states that (f(x)g(x))′=f(x)′g(x)+f(x)g(x)′{\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,
3x2+3y2dydx=3a(xdydx+y) ⇒3x2+3y2dydx=3axdydx+3ay  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 dydx\dfrac{{dy}}{{dx}} to one side.
3y2dydx−3axdydx=3ay−3x2 ⇒(3y2−3ax)dydx=3ay−3x2  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(y2−ax)dydx=3(ay−x2) ⇒(y2−ax)dydx=(ay−x2)  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 (y2−ax)\left( {{y^2} - ax} \right)
Hence,
dydx=ay−x2y2−ax\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 yy in terms of xx.