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Find $\dfrac{{dy}}{{dx}}{\text{ of }}{x^3} + {x^2}y + x{y^2} + {y^3} = 81$

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Hint: Apply chain rule. Also differentiation of a constant term is always zero.

Given equation, ${x^3} + {x^2}y + x{y^2} + {y^3} = 81$
Differentiate both sides with respect to $x$, we get
$
  \dfrac{{d\left( {{x^3} + {x^2}y + x{y^2} + {y^3}} \right)}}{{dx}} = \dfrac{{d\left( {81} \right)}}{{dx}} \\
   \Rightarrow 3{x^2} + \left( {y.2x + {x^2}\dfrac{{dy}}{{dx}}} \right) + \left( {{y^2}.1 + x.2y.\dfrac{{dy}}{{dx}}} \right) + 3{y^2}\dfrac{{dy}}{{dx}} = 0 \\
$
Rearranging the above equation, we get
\[
   \Rightarrow \left( {{x^2} + 2xy + 3{y^2}} \right)\dfrac{{dy}}{{dx}} + \left( {3{x^2} + 2xy + {y^2}} \right) = 0 \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \left( {3{x^2} + 2xy + {y^2}} \right)}}{{\left( {{x^2} + 2xy + 3{y^2}} \right)}} \\
 \]

Note: Whenever there are two different functions inside a derivative, the first step you'll need to take is to use the product rule. This rule tells you what to do when you are trying to take the derivative of the product of two functions. The product rule says that if you have two functions f and g, then the derivative of fg is fg' + f'g.



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