Question

How do you implicitly differentiate $1 = - xy + x + y - {y^2} + {x^2}$ ?

Answer 1

Hint : Start by considering $f(x)$ as the function of $x$ . Remember that the variable $y$ is the name of some function of $x$ which is not known to us . Apply the differentiation on both sides of the equation .Apply the chain rule for the terms containing product of $x\,and\,y$ to solve the derivative. We will solve the derivatives of each term separately. Now collect all the terms containing $\dfrac{{dy}}{{dx}}$ on one side and solve the equation for $\dfrac{{dy}}{{dx}}$ to get the required result.
Formula:
$\dfrac{d}{{dx}}\left( x \right) = 1 \\\ \dfrac{d}{{dx}}\left( 1 \right) = 0 \\\ \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} \;$

Complete step-by-step answer :
We are given $1 = - xy + x + y - {y^2} + {x^2}$ and we have to find its implicit differentiation with respect to $x$ .
$\Rightarrow 1 = - xy + x + y - {y^2} + {x^2}$
Rewriting by Reversing the equation,
$\Rightarrow - xy + x + y - {y^2} + {x^2} = 1$
Differentiating both sides of the equation with respect to $x$ .
$\Rightarrow \dfrac{d}{{dx}}\left( { - xy + x + y - {y^2} + {x^2}} \right) = \dfrac{d}{{dx}}\left( 1 \right)$
Remember one thing the $y$ in the above equation is the name of some other function of $x$ ,which we don’t know. $- xy$ is actually $- x\left( {some\,function\,of\,x} \right)$
So, to differentiate $- xy$ , we will use the product and chain rules. Similarly, with the term ${y^2}$ which is actually ${\left( {some\,f\,of\,x} \right)^2}$
Product Rule can be expressed as : ${\left( {FS} \right)^\prime } = F'S + FS'$
Splitting the derivative into all the terms and applying product rule for $- xy$ term, we get the equation as
$\Rightarrow \left( {\dfrac{d}{{dx}}\left( { - x} \right).y + \left( { - x} \right)\dfrac{d}{{dx}}\left( y \right)} \right) + \dfrac{d}{{dx}}\left( x \right) + \dfrac{{dy}}{{dx}}\left( y \right) - \dfrac{d}{{dx}}\left( {{y^2}} \right) + \dfrac{d}{{dx}}\left( {{x^2}} \right) = \dfrac{d}{{dx}}\left( 1 \right)$
Now with the use of standard rules and properties of differentiation, we obtain the above equation as
$\Rightarrow \left( { - y - x\dfrac{{dy}}{{dx}}} \right) + 1 + \dfrac{{dy}}{{dx}} - 2y\dfrac{{dy}}{{dx}} + 2x = 0 \\\ \Rightarrow - y - x\dfrac{{dy}}{{dx}} + 1 + \dfrac{{dy}}{{dx}} - 2y\dfrac{{dy}}{{dx}} + 2x = 0 \;$
Now transposing all the terms other than containing $\dfrac{{dy}}{{dx}}$ from Left-hand side towards Right-Hand side
$\Rightarrow - x\dfrac{{dy}}{{dx}} + \dfrac{{dy}}{{dx}} - 2y\dfrac{{dy}}{{dx}} = y - 1 - 2x$
Pulling out common $\dfrac{{dy}}{{dx}}$ from the Left-hand side part of equation , we get

$$\Rightarrow \dfrac{{dy}}{{dx}}\left( { - x + 1 - 2y} \right) = y - 1 - 2x \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {y - 1 - 2x} \right)}}{{\left( { - x + 1 - 2y} \right)}} \;$$

Therefore, the implicit differentiation of the equation $1 = - xy + x + y - {y^2} + {x^2}$ is $\dfrac{{dy}}{{dx}} = \dfrac{{\left( {y - 1 - 2x} \right)}}{{\left( { - x + 1 - 2y} \right)}}$.
So, the correct answer is “$\dfrac{{dy}}{{dx}} = \dfrac{{\left( {y - 1 - 2x} \right)}}{{\left( { - x + 1 - 2y} \right)}}$”.

Note : A derivative is the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations. In general, scientists observe changing systems to obtain rate of change of some variable of interest, incorporate this information into some differential equation, and use integration techniques to obtain a function that can be used to predict the behaviour of the original system under diverse conditions.