Question

How do you find $\dfrac{{dy}}{{dx}}$ by implicit differentiation given ${x^2} + 3xy + {y^2} = 0$ ?

Answer 1

In the given question, we have to differentiate the given equation with respect to x.
We use the rules of derivative functions to derive the given equation. For the first term we make use of the power rule of differentiation. Since the middle term is a product of two variables, we use product rule to simplify it. To the last term we simplify it using chain rule of differentiation. So after applying all the rules mentioned above, we try to get $\dfrac{{dy}}{{dx}}$.
Complete step by step solution:
Given the equation of the form,
${x^2} + 3xy + {y^2} = 0$ …… (1)
Note that the above equation is a quadratic equation in two variables x and y.
We are asked to find $\dfrac{{dy}}{{dx}}$ for the above equation.
Differentiating the above equation (1) with respect to x, we have,
$\dfrac{d}{{dx}}({x^2} + 3xy + {y^2}) = \dfrac{{d(0)}}{{dx}}$ …… (2)
We differentiate the terms one by one with respect to x.
Let us consider the first term which is ${x^2}$. And we find $\dfrac{d}{{dx}}({x^2})$
Note that it is of the form ${x^n}$. We have the power rule of differentiation given by,
$\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$.
Here $n = 2$.
Hence we get,
$\dfrac{d}{{dx}}({x^2}) = 2{x^{2 - 1}}$
$\Rightarrow \dfrac{d}{{dx}}({x^2}) = 2x$
Now we differentiate the middle term which is $3xy$with respect to x.
Here we differentiate $xy$ using the product rule of differentiation and multiply the obtained answer by 3.
$\dfrac{d}{{dx}}(xy) = \dfrac{{dx}}{{dx}} \cdot y + \dfrac{{dy}}{{dx}} \cdot x$
$\Rightarrow \dfrac{d}{{dx}}(xy) = 1 \cdot y + x \cdot \dfrac{{dy}}{{dx}}$
$\Rightarrow \dfrac{d}{{dx}}(xy) = y + x \cdot \dfrac{{dy}}{{dx}}$
$\therefore \dfrac{d}{{dx}}(3xy) = 3\left( {y + x \cdot \dfrac{{dy}}{{dx}}} \right)$
$\Rightarrow \dfrac{d}{{dx}}(3xy) = 3y + 3x\dfrac{{dy}}{{dx}}$
Lastly, we differentiate ${y^2}$ with respect to x using the chain rule of differentiation.
$\dfrac{d}{{dx}}({y^2}) = 2y\dfrac{{dy}}{{dx}}$
We know that differentiation of the constant term is zero.
Hence, $\dfrac{{d(0)}}{{dx}} = 0$
Substituting all this in the equation (2), we get,
$\dfrac{d}{{dx}}({x^2}) + \dfrac{d}{{dx}}(3xy) + \dfrac{d}{{dx}}({y^2}) = \dfrac{{d(0)}}{{dx}}$
$\Rightarrow 2x + 3y + 3x\dfrac{{dy}}{{dx}} + 2y\dfrac{{dy}}{{dx}} = 0$
Combining the like terms we get,
$\Rightarrow 2x + 3y + \dfrac{{dy}}{{dx}}(3x + 2y) = 0$
Taking the terms which do not contain $\dfrac{{dy}}{{dx}}$ to the other side, we get,
$\Rightarrow \dfrac{{dy}}{{dx}}(3x + 2y) = - (2x + 3y)$
Transferring $3x + 2y$ to the other side of the equation we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{2x + 3y}}{{3x + 2y}}$

Therefore, implicit differentiation of the equation ${x^2} + 3xy + {y^2} = 0$ is given by,
$\dfrac{{dy}}{{dx}} = - \dfrac{{2x + 3y}}{{3x + 2y}}$.

Note: The process to find the derivative of the function is called differentiation. In differentiation, there is an instantaneous rate of change of the functions based on the variable.
Implicit differentiation is a way of differentiation where we have a function in terms of both x and y. In implicit differentiation each side of the equation with the two variables is differentiated. This is done by treating one of the variables as the function of the other.