Question

How do you differentiate $x + y = xy$?

Answer 1

Derivative are defined as the varying rate of a function with respect to an independent variable. To differentiate the right hand side of the equation we use the product rule. That is if we have $y = uv$ then its differentiation with respect to ‘x’ is $\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}$. We solve this using implicit differentiation. We know that the differentiation of ‘x’ with respect to ‘x’ is 1.

Complete step-by-step solution:
Given, $x + y = xy$.
Now differentiate implicitly with respect to ‘x’.
$\dfrac{d}{{dx}}\left( {x + y} \right) = \dfrac{d}{{dx}}\left( {xy} \right)$
Applying the product rule $\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}$ in the right hand side of the differential equation, Where $u = x$ and $v = y$, then
$\dfrac{d}{{dx}}(x) + \dfrac{d}{{dx}}(y) = x \times \dfrac{d}{{dx}}(y) + y\dfrac{d}{{dx}}(x)$
$1 + \dfrac{{dy}}{{dx}} = x.\dfrac{{dy}}{{dx}} + y$
Grouping $\dfrac{{dy}}{{dx}}$ on one side we have,
$\dfrac{{dy}}{{dx}} - x.\dfrac{{dy}}{{dx}} = y - 1$
Taking $\dfrac{{dy}}{{dx}}$ as common we have,
$\dfrac{{dy}}{{dx}}\left( {1 - x} \right) = y - 1$
Divide the whole differential equation by $(x - 1)$, we have,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y - 1}}{{1 - x}}$.
Thus the differentiation of $x + y = xy$ is $\dfrac{{y - 1}}{{1 - x}}$.

Note: We know the differentiation of ${x^n}$ with respect to ‘x’ is $\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}$. We know the differentiation of ${y^n}$ with respect to ‘x’ is $\dfrac{{d({y^n})}}{{dx}} = n.{y^{n - 1}}\dfrac{{dy}}{{dx}}$.

We also have different rules in the differentiation. Those are
i) Linear combination rule: The linearity law is very important to emphasize its nature with alternate notation. Symbolically it is specified as $h'(x) = af'(x) + bg'(x)$

ii) Quotient rule: The derivative of one function divided by other is found by quotient rule such as${\left[ {\dfrac{{f(x)}}{{g(x)}}} \right]’} = \dfrac{{g(x)f’(x) - f(x)g’(x)}}{{{{\left[ {g(x)} \right]}^2}}}$.

iii) Product rule: When a derivative of a product of two function is to be found, then we use product rule that is $\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}$.

iv) Chain rule: To find the derivative of composition function or function of a function, we use chain rule. That is $fog'({x_0}) = [(f'og)({x_0})]g'({x_0})$. We use these rules depending on the given problem.