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

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

Answer

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

Step-by-step explanation

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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

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Answer

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

Main topic:

Differential calculus

Used formulas:

5. See formulas

Time to solve it:

~ 1.93 seconds