What is the derivative of ((e^{b/x}))/((e^{b/x)-1)} ?

The derivative of ((e^{b/x}))/((e^{b/x)-1)} is (be^{b/x})/(x^2(e^{b/x)-1)^2}

What is the first derivative of ((e^{b/x}))/((e^{b/x)-1)} ?

The first derivative of ((e^{b/x}))/((e^{b/x)-1)} is (be^{b/x})/(x^2(e^{b/x)-1)^2}

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derivative of

\frac{( e ^{\frac{b}{x}} )}{( e ^{\frac{b}{x}} - 1 )}

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

Solution steps

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

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

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

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

Simplify

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

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

\frac{d}{d x} ((x^{2} sin (x))^{5})

\int \frac{e ^{x}}{x + 1} d x

\frac{d}{d t} (\frac{sec ^{2} ( t )}{sec ( t ) tan ( t )})

\int_{- 2}^{1} \frac{1}{x ^{4}} d x

\frac{\partial}{\partial x} (2 y^{2} + 3 x)

What is the derivative of ((e^{b/x}))/((e^{b/x)-1)} ?
The derivative of ((e^{b/x}))/((e^{b/x)-1)} is (be^{b/x})/(x^2(e^{b/x)-1)^2}
What is the first derivative of ((e^{b/x}))/((e^{b/x)-1)} ?
The first derivative of ((e^{b/x}))/((e^{b/x)-1)} is (be^{b/x})/(x^2(e^{b/x)-1)^2}