Question

How do you differentiate the expression ${{e}^{xy}}+y=x-1$?

Answer 1

From the question we are asked to differentiate the equation ${{e}^{xy}}+y=x-1$. So, to solve the question we will differentiate each term separately, and its formulae which we will discuss below. After differentiating each term separately we will put each term back into the equation and proceed the further calculation to solve for $\dfrac{dy}{dx}$.

Complete step-by-step solution:
Firstly, we will differentiate the equation on both sides.
By differentiating both sides we will get,
$\Rightarrow {{e}^{xy}}+y=x-1$
$\Rightarrow \dfrac{d}{dx}\left( {{e}^{xy}} \right)+\dfrac{dy}{dx}=\dfrac{dx}{dx}-\dfrac{d}{dx}\left( 1 \right)$
Now we have four terms in the equation.
Now we will differentiate each term separately, we will start from left side onwards.
Differentiation of first term,
Let $u=xy$ then,
$\Rightarrow \dfrac{du}{dx}=\dfrac{d\left( xy \right)}{dx}$
Using product rule,
$\Rightarrow \dfrac{d\left( xy \right)}{dx}=\dfrac{d\left( x \right)}{dx}y+x\dfrac{d\left( y \right)}{dx}$
$\Rightarrow \dfrac{du}{dx}=y+x\dfrac{dy}{dx}$
Let this is equation 1
Now, we will find the differentiation of $\left( {{e}^{xy}} \right)$, it can be done by using the chain rule.
Now by using the chain rule;
$\Rightarrow \dfrac{d}{dx}\left( {{e}^{xy}} \right)=\dfrac{d\left( {{e}^{u}} \right)}{du}\times \dfrac{du}{dx}$
$\Rightarrow \dfrac{d}{dx}\left( {{e}^{xy}} \right)=\left( {{e}^{u}} \right)\times \dfrac{du}{dx}$
As we already know that differentiation of exponential is same, and we will substitute the equation 1 here,
$\Rightarrow \dfrac{d}{dx}\left( {{e}^{xy}} \right)={{e}^{xy}}\left( y+x\dfrac{dy}{dx} \right)$
$\Rightarrow \dfrac{d}{dx}\left( {{e}^{xy}} \right)=y{{e}^{xy}}+x{{e}^{xy}}\dfrac{dy}{dx}$
Now we will move to the second term,
$\Rightarrow \dfrac{dy}{dx}$
$\Rightarrow \dfrac{d\left( y \right)}{dx}=\dfrac{dy}{dx}$
Now, we will move to the third term,
$\Rightarrow \dfrac{dx}{dx}$
$\Rightarrow \dfrac{dx}{dx}=1$
Now, we will move to the fourth term,
$\Rightarrow \dfrac{d}{dx}\left( 1 \right)$
$\Rightarrow \dfrac{d}{dx}\left( 1 \right)=0$
Now, we will put the terms back into the equation,
By putting terms back into the equation, we will get,
$\Rightarrow \dfrac{d}{dx}\left( {{e}^{xy}} \right)+\dfrac{dy}{dx}=\dfrac{dx}{dx}-\dfrac{d}{dx}\left( 1 \right)$
$\Rightarrow y{{e}^{xy}}+x{{e}^{xy}}\dfrac{dy}{dx}+\dfrac{dy}{dx}=1$
Now, we have to take the common from the left-hand side and we will send remaining part to the right-hand side,
$\Rightarrow \dfrac{dy}{dx}\left( x{{e}^{xy}}+1 \right)=1-y{{e}^{xy}}$
by simplifying further, we will get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1-y{{e}^{xy}}}{\left( x{{e}^{xy}}+1 \right)}$
Therefore, the differentiation of ${{e}^{xy}}+y=x-1$ is $\dfrac{dy}{dx}=\dfrac{1-y{{e}^{xy}}}{\left( x{{e}^{xy}}+1 \right)}$.

Note: Students should know the basic rules of differentiation like product rule or UV rule, quotient rule, chain rule etc. Students should be careful while doing calculations and signs. Students should strike the idea while reading the problem. They will get the ideas by practicing more problems like this by practicing only they will get the ideas. Students should recall all the basic formulas of the differentiation.