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

The derivative of 1/([e^{c/x)-1]} is (ce^{c/x})/(x^2(e^{c/x)-1)^2}

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

The first derivative of 1/([e^{c/x)-1]} is (ce^{c/x})/(x^2(e^{c/x)-1)^2}

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derivative of

\frac{1}{[ e ^{\frac{c}{x}} - 1 ]}

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

Solution steps

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{1}{( x + 2 )}

2 x - lo g_{10} (x) - 7

\frac{s}{( s ^{2} + 1 ) ^{2}}

8 x^{- 1}

2 x^{3} + 3 lo g_{10} (x) + 2 sin (x)

What is the derivative of 1/([e^{c/x)-1]} ?
The derivative of 1/([e^{c/x)-1]} is (ce^{c/x})/(x^2(e^{c/x)-1)^2}
What is the first derivative of 1/([e^{c/x)-1]} ?
The first derivative of 1/([e^{c/x)-1]} is (ce^{c/x})/(x^2(e^{c/x)-1)^2}