Question

If $xy\log \left( x+y \right)=1$, prove that $\dfrac{dy}{dx}=-\dfrac{y\left( {{x}^{2}}y+x+y \right)}{x\left( x{{y}^{2}}+x+y \right)}$.

Answer 1

In the given question, we are given a trigonometric expression which we have to use in order to prove the given expression. We will first differentiate the given trigonometric function, $xy\log \left( x+y \right)=1$. After differentiating the expression, we will separate out the terms with and without $\dfrac{dy}{dx}$. Then, we will modify the obtained expression such that it appears similar to the expression asked in the question. Hence, we will have proved the required expression.

Complete step by step answer:
According to the given question, we are given a trigonometric expression which we have to use in order to prove another expression.
We have,
$xy\log \left( x+y \right)=1$----(1)
We will now differentiate both the sides of the equation (1). We will now consider $xy$ as the first term and $\log \left( x+y \right)$ as the second term. Applying the product rule, which is $\dfrac{d}{dx}\left( I.II \right)=I\dfrac{d}{dx}(II)+II\dfrac{d}{dx}(I)$, we get,
$\Rightarrow xy\dfrac{d}{dx}\left( \log \left( x+y \right) \right)+\log \left( x+y \right)\dfrac{d}{dx}\left( xy \right)=\dfrac{d}{dx}(1)$
We will write the derivative of logarithm function as well as $xy$, on which product rule will be applied, and then the derivative of a constant, which is, 1. We have,
$\Rightarrow xy\left( \dfrac{1}{\left( x+y \right)}\left( 1+\dfrac{dy}{dx} \right) \right)+\log \left( x+y \right)\left( x\dfrac{dy}{dx}+y \right)=0$
$\Rightarrow \dfrac{xy}{\left( x+y \right)}\left( 1+\dfrac{dy}{dx} \right)+\log \left( x+y \right)\left( x\dfrac{dy}{dx}+y \right)=0$
We will now open up the brackets and write the product of the respective terms and we get,
$\Rightarrow \dfrac{xy}{\left( x+y \right)}+\dfrac{xy}{\left( x+y \right)}\dfrac{dy}{dx}+x\log \left( x+y \right)\dfrac{dy}{dx}+y\log \left( x+y \right)=0$
From the given trigonometric expression that we have, which is, $xy\log \left( x+y \right)=1$, we can rewrite it as, $\log \left( x+y \right)=\dfrac{1}{xy}$. We get,
$\Rightarrow \dfrac{xy}{\left( x+y \right)}+\dfrac{xy}{\left( x+y \right)}\dfrac{dy}{dx}+x\left( \dfrac{1}{xy} \right)\dfrac{dy}{dx}+y\left( \dfrac{1}{xy} \right)=0$
Now, cancelling out the common terms, we get,
$\Rightarrow \dfrac{xy}{\left( x+y \right)}+\dfrac{xy}{\left( x+y \right)}\dfrac{dy}{dx}+\dfrac{1}{y}\dfrac{dy}{dx}+\dfrac{1}{x}=0$
Separating out the terms with $\dfrac{dy}{dx}$, we get,
$\Rightarrow \dfrac{xy}{\left( x+y \right)}\dfrac{dy}{dx}+\dfrac{1}{y}\dfrac{dy}{dx}=-\dfrac{1}{x}-\dfrac{xy}{\left( x+y \right)}$
$\Rightarrow \left( \dfrac{xy}{\left( x+y \right)}+\dfrac{1}{y} \right)\dfrac{dy}{dx}=-\left( \dfrac{1}{x}+\dfrac{xy}{\left( x+y \right)} \right)$
Taking the LCM on either side so that further computation can be done, we get,
$\Rightarrow \left( \dfrac{x{{y}^{2}}+x+y}{\left( x+y \right)y} \right)\dfrac{dy}{dx}=-\left( \dfrac{{{x}^{2}}y+x+y}{x\left( x+y \right)} \right)$
Cancelling out the similar terms across the equality, we get,
$\Rightarrow \left( \dfrac{x{{y}^{2}}+x+y}{y} \right)\dfrac{dy}{dx}=-\left( \dfrac{{{x}^{2}}y+x+y}{x} \right)$
We will now do the cross multiplication and write the expression as per asked in the question, so we get,
$\Rightarrow \dfrac{dy}{dx}=-\dfrac{y\left( {{x}^{2}}y+x+y \right)}{x\left( x{{y}^{2}}+x+y \right)}$
Hence, proved.

Note: While solving the expression, we need to have an eye on the expression we have to prove so that the obtained expression can be modified appropriately. The given expression, $xy\log \left( x+y \right)=1$, is applied in the above solution so as to make the expression void of logarithmic function. Also, the solution should be carried out step wise.