Question

If $x = y\log (xy)$ then find $\dfrac{{dy}}{{dx}}$ .

Answer 1

Hint:
We will use chain rule in derivatives that is $\dfrac{d}{{dx}}\left( {f(x)g(x)} \right) = g(x)\dfrac{{d(f(x))}}{{dx}} + f(x)\dfrac{{d(g(x))}}{{dx}}$
And use the fact that $\log (ab) = \log (a) + \log (b)$ ,and then we will separate the $\dfrac{{dy}}{{dx}}$ term to find the required expression.

Complete step by step solution:
Our main equation is
 $ \Rightarrow x = y\log (xy)$ …(i)
First, we will find the derivative of the expression $\log (xy)$ which is
 $ \Rightarrow \dfrac{{d\left( {\log (xy)} \right)}}{{dx}}$
Using $\log (ab) = \log (a) + \log (b)$ where $a = x,b = y$ we get
 $ \Rightarrow \dfrac{{d\left( {\log (x) + \log (y)} \right)}}{{dx}}$
Using \[\dfrac{{d(a + b)}}{{dx}} = \dfrac{{da}}{{dx}} + \dfrac{{db}}{{dx}}\] , we get,
 $ \Rightarrow \dfrac{{d(\log (x))}}{{dx}} + \dfrac{{d(\log (y))}}{{dx}}$
Using \[\dfrac{{d(\log x)}}{{dx}} = \dfrac{1}{x}\] , we get,
 $ \Rightarrow \dfrac{1}{x} + \dfrac{1}{y}\dfrac{{dy}}{{dx}}$ …(ii)
Now we will use our main equation (i) which is
 $ \Rightarrow x = y\log (xy)$
Now we will differentiate both side with respect to x, then we get
 $ \Rightarrow \dfrac{{dx}}{{dx}} = \dfrac{{d(y\log (xy))}}{{dx}}$
Using the chain rule in differentiating we get
 $ \Rightarrow 1 = \log (xy)\dfrac{{dy}}{{dx}} + y\dfrac{{d(\log (xy))}}{{dx}}$
Replacing $\dfrac{{d(\log (xy))}}{{dx}}$ by equation (ii) we get
 $ \Rightarrow 1 = \log (xy)\dfrac{{dy}}{{dx}} + y\left( {\dfrac{1}{x} + \dfrac{1}{y}\dfrac{{dy}}{{dx}}} \right)$
Separating $\dfrac{{dy}}{{dx}}$ we get
 \[ \Rightarrow 1 = \dfrac{y}{x} + \left( {\log (xy) + 1} \right)\dfrac{{dy}}{{dx}}\]
On rearranging we get,
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {1 - \dfrac{y}{x}} \right)}}{{\left( {\log (xy) + 1} \right)}}$
On multiplying and dividing by x we get,
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {x - y} \right)}}{{x\left( {\log (xy) + 1} \right)}}$
Which is our required expression.

Note:
Sometimes if we just simplify the main equation, differentiation just becomes easier. One thing that we must remember when differentiation becomes complicated is that we take a subpart of the equation and differentiate it so that when applying chain rule we just replace that part with its differentiation.