# Step-by-step Solution

## Find the derivative using the quotient rule $\frac{d}{dx}\left(\frac{e^{\left(x+y\right)}}{2y-e^{\left(x+y\right)}}\right)$

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### Videos

$\frac{e^{\left(x+y\right)}\left(2y-e^{\left(x+y\right)}\right)+e^{2\left(x+y\right)}}{\left(2y-e^{\left(x+y\right)}\right)^2}$

## Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(\frac{e^{\left(x+y\right)}}{2y-1\cdot e^{\left(x+y\right)}}\right)$
1

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{\frac{d}{dx}\left(e^{\left(x+y\right)}\right)\left(2y-e^{\left(x+y\right)}\right)-e^{\left(x+y\right)}\cdot\frac{d}{dx}\left(2y-e^{\left(x+y\right)}\right)}{\left(2y-e^{\left(x+y\right)}\right)^2}$
2

Applying the derivative of the exponential function

$\frac{1e^{\left(x+y\right)}\cdot\frac{d}{dx}\left(x+y\right)\left(2y-e^{\left(x+y\right)}\right)-e^{\left(x+y\right)}\cdot\frac{d}{dx}\left(2y-e^{\left(x+y\right)}\right)}{\left(2y-e^{\left(x+y\right)}\right)^2}$

$\frac{e^{\left(x+y\right)}\left(2y-e^{\left(x+y\right)}\right)+e^{2\left(x+y\right)}}{\left(2y-e^{\left(x+y\right)}\right)^2}$
$\frac{d}{dx}\left(\frac{e^{\left(x+y\right)}}{2y-1\cdot e^{\left(x+y\right)}}\right)$

Quotient rule

~ 0.86 seconds

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