Question

How do you differentiate $\ln \left( {\dfrac{y}{x}} \right) = xy$ ?

Answer 1

Hint: In the given problem, we are required to differentiate both sides of the equation $\ln \left( {\dfrac{y}{x}} \right) = xy$ with respect to x. Since, $\ln \left( {\dfrac{y}{x}} \right) = xy$ is an implicit function, so we will have to differentiate the function with the implicit method of differentiation. So, differentiation of $\ln \left( {\dfrac{y}{x}} \right) = xy$ with respect to x will be done layer by layer using the chain rule of differentiation as in the given function, we cannot isolate the variables $x$ and $y$.

Complete step by step answer:
Consider, $\ln \left( {\dfrac{y}{x}} \right) = xy$. Differentiating both sides of the equation with respect to x, we get,
$\dfrac{d}{{dx}}\left[ {\ln \left( {\dfrac{y}{x}} \right)} \right] = \dfrac{d}{{dx}}\left( {xy} \right)$
We know the product rule of differentiation as $\dfrac{d}{{dx}}\left( {f(x) \times g(x)} \right) = f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right) + g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right)$ .

So, we get the right side of the equation as,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {\ln \left( {\dfrac{y}{x}} \right)} \right] = x\dfrac{{dy}}{{dx}} + y\dfrac{{d\left( x \right)}}{{dx}}$
Now, we make use of the power rule of differentiation as $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$. So, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {\ln \left( {\dfrac{y}{x}} \right)} \right] = x\dfrac{{dy}}{{dx}} + y$
We consider $\dfrac{y}{x}$ as t. So, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {\ln t} \right] = x\dfrac{{dy}}{{dx}} + y$

Now, we know that the derivative of the logarithmic function $\ln x$ with respect to x is $\left( {\dfrac{1}{x}} \right)$. Hence, we get,
$ \Rightarrow \dfrac{1}{t}\left( {\dfrac{{dt}}{{dx}}} \right) = x\dfrac{{dy}}{{dx}} + y$
Now, putting t as $\left( {\dfrac{y}{x}} \right)$, we get,
$ \Rightarrow \dfrac{1}{{\left( {\dfrac{y}{x}} \right)}}\left( {\dfrac{{d\left( {\dfrac{y}{x}} \right)}}{{dx}}} \right) = x\dfrac{{dy}}{{dx}} + y$
Now, we use the quotient rule of differentiation as $\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right) - f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$ .
So, we get,
$ \Rightarrow \left( {\dfrac{x}{y}} \right)\left( {\dfrac{{x\dfrac{{dy}}{{dx}} - y\left( {\dfrac{{dx}}{{dx}}} \right)}}{{{x^2}}}} \right) = x\dfrac{{dy}}{{dx}} + y$

Opening the brackets,
$ \Rightarrow \left( {\dfrac{1}{y}} \right)\left( {\dfrac{{x\dfrac{{dy}}{{dx}} - y}}{x}} \right) = x\dfrac{{dy}}{{dx}} + y$
$ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} - \dfrac{1}{x} = x\dfrac{{dy}}{{dx}} + y$
Now, isolating the $\dfrac{{dy}}{{dx}}$ terms, we get,
$ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} - x\dfrac{{dy}}{{dx}} = \dfrac{1}{x} + y$
Taking $\dfrac{{dy}}{{dx}}$ common in left side of equation, we get,
$ \therefore \dfrac{{dy}}{{dx}} = \left( {\dfrac{{1 + xy}}{x}} \right)\left( {\dfrac{y}{{1 - xy}}} \right)$

Therefore, we get $\dfrac{{dy}}{{dx}}$ as $\left( {\dfrac{{1 + xy}}{x}} \right)\left( {\dfrac{y}{{1 - xy}}} \right)$.

Note: Implicit functions are those functions that involve two variables and the two variables are not separable and cannot be isolated from each other. Quotient rule of differentiation $\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right) - f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$ is used to find derivative of rational expressions whereas product rule helps in finding derivative of product functions.