Solveeit Logo
Login

Question

Question: How do you use implicit differentiation to find \[\dfrac{{dy}}{{dx}}\] given \[{x^2} + xy + {y^2} = ...

How do you use implicit differentiation to find dydx\dfrac{{dy}}{{dx}} given x2+xy+y2=9{x^2} + xy + {y^2} = 9 ?

Explanation

Solution

Here we differentiate the complete equation and apply the product rule of differentiation in LHS wherever required. Collect the values having dydx\dfrac{{dy}}{{dx}}and calculate its value by taking all other values to the opposite side of the equation.

  • Differentiation of xn{x^n} with respect to x is given by ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}
  • Product rule of differentiation: ddx(m×n)=m×dndx+n×dmdx\dfrac{d}{{dx}}(m \times n) = m \times \dfrac{{dn}}{{dx}} + n \times \dfrac{{dm}}{{dx}}

Complete step by step solution:
We are given the equation x2+xy+y2=9{x^2} + xy + {y^2} = 9
We differentiate both sides of the equation with respect to ‘x’.
⇒ddx(x2+xy+y2)=ddx(9)\Rightarrow \dfrac{d}{{dx}}\left( {{x^2} + xy + {y^2}} \right) = \dfrac{d}{{dx}}\left( 9 \right)
Break the differentiation in LHS of the equation
⇒ddx(x2)+ddx(xy)+ddx(y2)=ddx(9)\Rightarrow \dfrac{d}{{dx}}({x^2}) + \dfrac{d}{{dx}}(xy) + \dfrac{d}{{dx}}({y^2}) = \dfrac{d}{{dx}}\left( 9 \right)
Use the differentiation method to write differentiation of each term in LHS and apply product rule to the middle term in LHS of the equation
⇒2x2−1+(x×dydx+y×dxdx)+2y2−1dydx=0\Rightarrow 2{x^{2 - 1}} + \left( {x \times \dfrac{{dy}}{{dx}} + y \times \dfrac{{dx}}{{dx}}} \right) + 2{y^{2 - 1}}\dfrac{{dy}}{{dx}} = 0
Calculate the values in powers and cancel same terms from numerator and denominator in fraction
⇒2x+(xdydx+y)+2ydydx=0\Rightarrow 2x + \left( {x\dfrac{{dy}}{{dx}} + y} \right) + 2y\dfrac{{dy}}{{dx}} = 0
Collect the terms having dydx\dfrac{{dy}}{{dx}}common
⇒2x+y+(xdydx+2ydydx)=0\Rightarrow 2x + y + \left( {x\dfrac{{dy}}{{dx}} + 2y\dfrac{{dy}}{{dx}}} \right) = 0
Shift all other values except the bracket to RHS of the equation
⇒(xdydx+2ydydx)=−2x−y\Rightarrow \left( {x\dfrac{{dy}}{{dx}} + 2y\dfrac{{dy}}{{dx}}} \right) = - 2x - y
Take dydx\dfrac{{dy}}{{dx}}common in LHS of the equation
⇒dydx(x+2y)=−2x−y\Rightarrow \dfrac{{dy}}{{dx}}\left( {x + 2y} \right) = - 2x - y
Divide both sides of the equation by (x+2y)(x + 2y)
⇒dydx(x+2y)(x+2y)=−2x−y(x+2y)\Rightarrow \dfrac{{\dfrac{{dy}}{{dx}}\left( {x + 2y} \right)}}{{(x + 2y)}} = \dfrac{{ - 2x - y}}{{(x + 2y)}}
Cancel same terms from numerator and denominator in LHS of the equation
⇒dydx=−(2x+y)(x+2y)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - (2x + y)}}{{(x + 2y)}}
∴\therefore The value of dydx=−(2x+y)(x+2y)\dfrac{{dy}}{{dx}} = \dfrac{{ - (2x + y)}}{{(x + 2y)}}when x2+xy+y2=9{x^2} + xy + {y^2} = 9

Note: Students many times make mistakes when they forget to write the differentiation of ‘y’ with respect to ‘x’ after they have performed differentiation of the function containing ‘y’ with respect to ‘y’. Keep in mind we will get the valuedydx\dfrac{{dy}}{{dx}}only when we differentiate ‘y’ with respect to ‘x’. Also, many students make mistakes when shifting values from one side of the equation to another, keep in mind we always change sign from positive to negative and vice-versa when shifting values to the opposite side of the equation.