What is the derivative of (e^{t/2}-e)^{-1/3} ?

The derivative of (e^{t/2}-e)^{-1/3} is -(e^{t/2})/(6(e^{t/2)-e)^{4/3}}

What is the first derivative of (e^{t/2}-e)^{-1/3} ?

The first derivative of (e^{t/2}-e)^{-1/3} is -(e^{t/2})/(6(e^{t/2)-e)^{4/3}}

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derivative of

(e^{\frac{t}{2}} - e)^{- \frac{1}{3}}

- \frac{e ^{\frac{t}{2}}}{6 ( e ^{\frac{t}{2}} - e ) ^{\frac{4}{3}}}

Solution steps

\frac{d}{d t} ((e^{\frac{t}{2}} - e)^{- \frac{1}{3}})

Apply the chain rule:

- \frac{1}{3 ( e ^{\frac{t}{2}} - e ) ^{\frac{4}{3}}} \frac{d}{d t} (e^{\frac{t}{2}} - e)

= - \frac{1}{3 ( e ^{\frac{t}{2}} - e ) ^{\frac{4}{3}}} \frac{d}{d t} (e^{\frac{t}{2}} - e)

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

= - \frac{1}{3 ( e ^{\frac{t}{2}} - e ) ^{\frac{4}{3}}} \cdot \frac{1}{2} e^{\frac{t}{2}}

Simplify

- \frac{1}{3 ( e ^{\frac{t}{2}} - e ) ^{\frac{4}{3}}} \cdot \frac{1}{2} e^{\frac{t}{2}} : - \frac{e ^{\frac{t}{2}}}{6 ( e ^{\frac{t}{2}} - e ) ^{\frac{4}{3}}}

= - \frac{e ^{\frac{t}{2}}}{6 ( e ^{\frac{t}{2}} - e ) ^{\frac{4}{3}}}

\frac{\partial}{\partial x} (\frac{e ^{y}}{x + y ^{4}})

lo g_{b} (3)

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

y = - 6 x^{5} + 2

\frac{x}{e ^{x} - 1}

What is the derivative of (e^{t/2}-e)^{-1/3} ?
The derivative of (e^{t/2}-e)^{-1/3} is -(e^{t/2})/(6(e^{t/2)-e)^{4/3}}
What is the first derivative of (e^{t/2}-e)^{-1/3} ?
The first derivative of (e^{t/2}-e)^{-1/3} is -(e^{t/2})/(6(e^{t/2)-e)^{4/3}}