# Derive the function (e^(x+y))/(2y-1e^(x+y)) with respect to x

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

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

## Step by step solution

Problem

$\frac{d}{dx}\left(\frac{e^{\left(x+y\right)}}{2y-1\cdot e^{\left(x+y\right)}}\right)$
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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

Applying the derivative of the exponential function

$\frac{1\left(2y-e^{\left(y+x\right)}\right)\frac{d}{dx}\left(y+x\right)e^{\left(y+x\right)}-\frac{d}{dx}\left(2y-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
3

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{1\left(2y-e^{\left(y+x\right)}\right)\left(\frac{d}{dx}\left(y\right)+\frac{d}{dx}\left(x\right)\right)e^{\left(y+x\right)}-\left(\frac{d}{dx}\left(-e^{\left(y+x\right)}\right)+\frac{d}{dx}\left(2y\right)\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
4

The derivative of the constant function is equal to zero

$\frac{1\left(2y-e^{\left(y+x\right)}\right)\left(0+\frac{d}{dx}\left(x\right)\right)e^{\left(y+x\right)}-\left(\frac{d}{dx}\left(-e^{\left(y+x\right)}\right)+0\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
5

The derivative of the linear function is equal to $1$

$\frac{\left(0+1\right)\cdot 1\left(2y-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}-\left(\frac{d}{dx}\left(-e^{\left(y+x\right)}\right)+0\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
6

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$\frac{\left(0+1\right)\cdot 1\left(2y-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}-\left(0-\frac{d}{dx}\left(e^{\left(y+x\right)}\right)\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
7

Applying the derivative of the exponential function

$\frac{\left(0+1\right)\cdot 1\left(2y-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}-\left(0-1\cdot 1\frac{d}{dx}\left(y+x\right)e^{\left(y+x\right)}\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
8

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{\left(0+1\right)\cdot 1\left(2y-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}-\left(0-1\cdot 1\left(\frac{d}{dx}\left(y\right)+\frac{d}{dx}\left(x\right)\right)e^{\left(y+x\right)}\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
9

The derivative of the constant function is equal to zero

$\frac{\left(0+1\right)\cdot 1\left(2y-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}-\left(0-1\cdot 1\left(0+\frac{d}{dx}\left(x\right)\right)e^{\left(y+x\right)}\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
10

The derivative of the linear function is equal to $1$

$\frac{\left(0+1\right)\cdot 1\left(2y-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}-\left(0-1\cdot \left(0+1\right)\cdot 1e^{\left(y+x\right)}\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
11

Add the values $1$ and $0$

$\frac{1\cdot 1\left(2y-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}-\left(0-1\cdot 1\cdot 1e^{\left(y+x\right)}\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
12

Multiply $1$ times $1$

$\frac{1\left(2y-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}-\left(0-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
13

$x+0=x$, where $x$ is any expression

$\frac{1\left(2y-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}-1\left(-1\right)e^{\left(y+x\right)}e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
14

Multiply $-1$ times $-1$

$\frac{1e^{\left(y+x\right)}e^{\left(y+x\right)}+1\left(2y-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
15

Any expression multiplied by $1$ is equal to itself

$\frac{e^{\left(y+x\right)}e^{\left(y+x\right)}+\left(2y-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
16

When multiplying exponents with same base you can add the exponents

$\frac{\left(e^{\left(y+x\right)}\right)^2+\left(2y-e^{\left(y+x\right)}\right)e^{\left(y+x\right)}}{\left(2y-e^{\left(y+x\right)}\right)^2}$
17

Applying the power of a power property

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

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

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### Main topic:

Differential calculus

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