Question

How do you differentiate ${{e}^{y}}\sin x=x+xy$?

Answer 1

Hint: In this question, we need to differentiate an implicit function of x and y. For this, we will differentiate with respect to x. Here y is a function of x. So while differentiating terms having y we will differentiate them as usual, then multiply it with $\dfrac{dy}{dx}$ as per the chain rule. At last we will rearrange the equation such that we have $\dfrac{dy}{dx}$ on one side and the rest of the terms on the other side. We will use following formulas of derivatives,
(I) Derivative of ${{e}^{x}}$ is ${{e}^{x}}$.
(II) Derivative of sinx is cosx.
(III) Product rule between two functions u and v is given by $\dfrac{d}{dx}\left( u\cdot v \right)=u'v+v'u$.
(IV) $\dfrac{dx}{dx}=1$.

Complete step by step solution:
Here we are given the implicit equation of x and y as ${{e}^{y}}\sin x=x+xy\cdots \cdots \ \cdots \left( 1 \right)$.
We need to differentiate it so we need to find $\dfrac{dy}{dx}$.
Let us solve both sides separately.
Left side of the equation is ${{e}^{y}}\sin x$.
Since two terms are multiplied so we will use product rule according to which, for two function u and v $\dfrac{d}{dx}\left( u\cdot v \right)=u'v+v'u$ so, taking derivative with respect to x we get $\dfrac{{{e}^{y}}d\sin x}{dx}+\sin x\dfrac{d{{e}^{y}}}{dx}$.
As we know that, derivative of sinx is cosx so we get ${{e}^{y}}\cos x+\sin x\dfrac{d{{e}^{y}}}{dx}$.
We know the derivative of ${{e}^{x}}$ is ${{e}^{x}}$. Here y is function of x so derivative of ${{e}^{y}}$ will be ${{e}^{y}}\dfrac{dy}{dx}$ by chain rule, we get derivative of left side ${{e}^{y}}\cos x+\sin x\cdot {{e}^{y}}\dfrac{dy}{dx}\ldots \ldots \ldots \left( 2 \right)$.
Now, let us derive the right side. We have the right side as $x+xy$.
Taking derivatives with respect to x we get $\dfrac{d}{dx}\left( x+xy \right)$.
Separating derivatives for both terms we have $\dfrac{dx}{dx}+\dfrac{d}{dx}\left( xy \right)$.
We know that $\dfrac{dx}{dx}=1$ and let us use product rule on $\dfrac{dxy}{dx}$ we get,
$1+x\dfrac{dy}{dx}+y\dfrac{dx}{dx}\Rightarrow 1+x\dfrac{dy}{dx}+y$.
So the derivative of the right side $1+y+x\dfrac{dy}{dx}\cdots \cdots \cdots \left( 3 \right)$.
Now when we take derivative of (1) from (2) and (3) we get ${{e}^{y}}\left( \dfrac{dy}{dx} \right)\sin x+{{e}^{y}}\cos x=1+y+x\left( \dfrac{dy}{dx} \right)$.
Now let us take terms having $\dfrac{dy}{dx}$ on the left side of the equation and rest of the terms on the right side of the equation we get ${{e}^{y}}\sin x\left( \dfrac{dy}{dx} \right)-x\left( \dfrac{dy}{dx} \right)=1+y-{{e}^{y}}\cos x$.
Now let us take $\dfrac{dy}{dx}$ common from the left side of the equation we get $\dfrac{dy}{dx}\left( {{e}^{y}}\sin x-x \right)=1+y-{{e}^{y}}\cos x$.
We need to only $\dfrac{dy}{dx}$ at one side and rest of the terms on the other side so let us divide both sides of the equation we get $\dfrac{dy}{dx}\dfrac{\left( {{e}^{y}}\sin x-x \right)}{\left( {{e}^{y}}\sin x-x \right)}=\dfrac{1+y-{{e}^{y}}\cos x}{\left( {{e}^{y}}\sin x-x \right)}\Rightarrow \dfrac{dy}{dx}=\dfrac{1+y-{{e}^{y}}\cos x}{\left( {{e}^{y}}\sin x-x \right)}$.
This is our required derivative.

Note: Students should not forget to use chain rule while taking derivatives of terms containing y. Some students make the mistake of taking y as constant but here y is a function of x. Always try to give a final answer as $\dfrac{dy}{dx}$ only. Keep in mind the formula of derivation.