# Find \[\dfrac{{dy}}{{dx}}\] of \[{x^2} + xy + {y^2} = 100\].

Answer

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Hint:- Use the product rule to find \[\dfrac{{dy}}{{dx}}\] of \[xy\] the derivative.

Given equation in the question is ,

\[ \Rightarrow {x^2} + xy + {y^2} = 100\] (1)

We had to find the \[\dfrac{{dy}}{{dx}}\] of the given equation 1. So, for that,

We must find the derivative of the given equation 1 with respect to x.

Finding \[\dfrac{{dy}}{{dx}}\] of the given equation,

\[ \Rightarrow 2x + \left( {y + x\dfrac{{dy}}{{dx}}({\text{By applying product rule)}}} \right) + 2y\dfrac{{dy}}{{dx}} = 0\] (2)

Now solving equation 2.

Taking \[\dfrac{{dy}}{{dx}}\] common from equation 2. It becomes,

\[ \Rightarrow 2x + y + \dfrac{{dy}}{{dx}}(x + 2y) = 0\]

Now taking \[(2x + y)\] to the RHS of the above equation. It becomes,

\[ \Rightarrow \dfrac{{dy}}{{dx}}(x + 2y) = - \left( {2x + y} \right)\]

Now, dividing both sides of the above equation by \[(x + 2y)\]. We get,

\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\left( {2x + y} \right)}}{{(x + 2y)}}\]

Hence, value of \[\dfrac{{dy}}{{dx}}\] for the given equation will be \[ - \dfrac{{\left( {2x + y} \right)}}{{(x + 2y)}}\].

Note:- Whenever we came up with this type of problem then first, find derivative of given

equation with respect to x, using different derivative formulas and various rules like product

rule, quotient rule and chain rule etc. Then take all the terms with \[\dfrac{{dy}}{{dx}}\] to one side of the equation.

As this will be the easiest and efficient way to find the value of \[\dfrac{{dy}}{{dx}}\] for the given equation.

Given equation in the question is ,

\[ \Rightarrow {x^2} + xy + {y^2} = 100\] (1)

We had to find the \[\dfrac{{dy}}{{dx}}\] of the given equation 1. So, for that,

We must find the derivative of the given equation 1 with respect to x.

Finding \[\dfrac{{dy}}{{dx}}\] of the given equation,

\[ \Rightarrow 2x + \left( {y + x\dfrac{{dy}}{{dx}}({\text{By applying product rule)}}} \right) + 2y\dfrac{{dy}}{{dx}} = 0\] (2)

Now solving equation 2.

Taking \[\dfrac{{dy}}{{dx}}\] common from equation 2. It becomes,

\[ \Rightarrow 2x + y + \dfrac{{dy}}{{dx}}(x + 2y) = 0\]

Now taking \[(2x + y)\] to the RHS of the above equation. It becomes,

\[ \Rightarrow \dfrac{{dy}}{{dx}}(x + 2y) = - \left( {2x + y} \right)\]

Now, dividing both sides of the above equation by \[(x + 2y)\]. We get,

\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\left( {2x + y} \right)}}{{(x + 2y)}}\]

Hence, value of \[\dfrac{{dy}}{{dx}}\] for the given equation will be \[ - \dfrac{{\left( {2x + y} \right)}}{{(x + 2y)}}\].

Note:- Whenever we came up with this type of problem then first, find derivative of given

equation with respect to x, using different derivative formulas and various rules like product

rule, quotient rule and chain rule etc. Then take all the terms with \[\dfrac{{dy}}{{dx}}\] to one side of the equation.

As this will be the easiest and efficient way to find the value of \[\dfrac{{dy}}{{dx}}\] for the given equation.

Last updated date: 23rd Sep 2023

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