What is the derivative of y=(e^{x/2})/(x+1) ?

The derivative of y=(e^{x/2})/(x+1) is (e^{x/2}x-e^{x/2})/(2(x+1)^2)

What is the first derivative of y=(e^{x/2})/(x+1) ?

The first derivative of y=(e^{x/2})/(x+1) is (e^{x/2}x-e^{x/2})/(2(x+1)^2)

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derivative of

y = \frac{e ^{\frac{x}{2}}}{x + 1}

\frac{e ^{\frac{x}{2}} x - e ^{\frac{x}{2}}}{2 ( x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{e ^{\frac{x}{2}}}{x + 1})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( e ^{\frac{x}{2}} ) ( x + 1 ) - \frac{d}{d x} ( x + 1 ) e ^{\frac{x}{2}}}{( x + 1 ) ^{2}}

\frac{d}{d x} (e^{\frac{x}{2}}) = e^{\frac{x}{2}} \frac{1}{2}

\frac{d}{d x} (x + 1) = 1

= \frac{e ^{\frac{x}{2}} \frac{1}{2} ( x + 1 ) - 1 \cdot e ^{\frac{x}{2}}}{( x + 1 ) ^{2}}

Simplify

\frac{e ^{\frac{x}{2}} \frac{1}{2} ( x + 1 ) - 1 \cdot e ^{\frac{x}{2}}}{( x + 1 ) ^{2}} : \frac{e ^{\frac{x}{2}} x - e ^{\frac{x}{2}}}{2 ( x + 1 ) ^{2}}

= \frac{e ^{\frac{x}{2}} x - e ^{\frac{x}{2}}}{2 ( x + 1 ) ^{2}}

\frac{1}{x ^{6}}

\int_{- \infty}^{0} x e^{3 x} d x

\frac{d}{d x} (2 (x)^{3} ln (x))

\int x e^{z} + y^{2} d z

(\frac{x ^{2} + 3}{x ^{2} - 3})^{8}

What is the derivative of y=(e^{x/2})/(x+1) ?
The derivative of y=(e^{x/2})/(x+1) is (e^{x/2}x-e^{x/2})/(2(x+1)^2)
What is the first derivative of y=(e^{x/2})/(x+1) ?
The first derivative of y=(e^{x/2})/(x+1) is (e^{x/2}x-e^{x/2})/(2(x+1)^2)