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Question

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

Find dydx\dfrac{{dy}}{{dx}} of x2+xy+y2=100{x^2} + xy + {y^2} = 100.

Explanation

Solution

Hint:- Use the product rule to find dydx\dfrac{{dy}}{{dx}} of xyxy the derivative.

Given equation in the question is ,
x2+xy+y2=100\Rightarrow {x^2} + xy + {y^2} = 100 (1)
We had to find the dydx\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 dydx\dfrac{{dy}}{{dx}} of the given equation,
2x+(y+xdydx(By applying product rule))+2ydydx=0\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 dydx\dfrac{{dy}}{{dx}} common from equation 2. It becomes,
2x+y+dydx(x+2y)=0\Rightarrow 2x + y + \dfrac{{dy}}{{dx}}(x + 2y) = 0
Now taking (2x+y)(2x + y) to the RHS of the above equation. It becomes,
dydx(x+2y)=(2x+y)\Rightarrow \dfrac{{dy}}{{dx}}(x + 2y) = - \left( {2x + y} \right)
Now, dividing both sides of the above equation by (x+2y)(x + 2y). We get,
dydx=(2x+y)(x+2y)\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\left( {2x + y} \right)}}{{(x + 2y)}}
Hence, value of dydx\dfrac{{dy}}{{dx}} for the given equation will be (2x+y)(x+2y) - \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 dydx\dfrac{{dy}}{{dx}} to one side of the equation.
As this will be the easiest and efficient way to find the value of dydx\dfrac{{dy}}{{dx}} for the given equation.