Question

Find the derivative of the following $x{{e}^{y}}+1=xy$

Answer 1

Hint: To solve the above problem we have to know the basic derivatives of functions. To solve further we have to use the uv rule. If u and v are two functions of x, then the derivative of the product uv is given by $\dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$.

Complete step-by-step answer:
Given $x{{e}^{y}}+1=xy$
Differentiating both sides of the equation we get,
$\dfrac{d}{dx}\left( x{{e}^{y}}+1 \right)=\dfrac{d}{dx}\left( xy \right)$
Now taking the L.H.S term,
$\dfrac{d}{dx}\left( x{{e}^{y}}+1 \right)$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
$\dfrac{d}{dx}\left( x \right)=1$ . . . . . . . . . . . . . . . . . . . . . . . . . (1)
$\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}$ . . . . . . . . . . . . . . . . . . . . . . . . (2)
By differentiating the above term (a) we get
$={{e}^{y}}+x{{e}^{y}}\dfrac{d}{dx}\left[ y \right]$
Now taking R.H.S term,
$\dfrac{d}{dx}\left( xy \right)$
We have seen hints how the uv rule works.
$\dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$
$=y+x\dfrac{d}{dx}\left[ y \right]$
Reforming the equation by setting the L.H.S =R.H.S
{{e}^{y}}+x{{e}^{y}}\dfrac{d}{dx}\left[ y \right]$$$$=y+x\dfrac{d}{dx}\left[ y \right]
Further solving for $\dfrac{dy}{dx}$
$x{{e}^{y}}\dfrac{d}{dx}\left[ y \right]-x\dfrac{d}{dx}\left[ y \right]=y-{{e}^{y}}$
$(x{{e}^{y}}-x)\dfrac{d}{dx}\left[ y \right]=y-{{e}^{y}}$
$\dfrac{dy}{dx}=\dfrac{y}{(x{{e}^{y}}-x)}-\dfrac{{{e}^{y}}}{(x{{e}^{y}}-x)}$

Note: In the above problem we have used the product rule. After using this rule we have got the values for $\dfrac{dy}{dx}$. Further solving for $\dfrac{dy}{dx}$made us towards a solution. If we are doing derivative means we are finding the slope of a function. Care should be taken while doing calculations.