What is the derivative of 20(1+e^{10-t})^{-1} ?

The derivative of 20(1+e^{10-t})^{-1} is (20e^{-t+10})/((1+e^{10-t))^2}

What is the first derivative of 20(1+e^{10-t})^{-1} ?

The first derivative of 20(1+e^{10-t})^{-1} is (20e^{-t+10})/((1+e^{10-t))^2}

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derivative of

20 (1 + e^{10 - t})^{- 1}

\frac{20 e ^{- t + 10}}{( 1 + e ^{10 - t} ) ^{2}}

Solution steps

\frac{d}{d t} (20 (1 + e^{10 - t})^{- 1})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 20 \frac{d}{d t} ((1 + e^{10 - t})^{- 1})

Apply the chain rule:

- \frac{1}{( 1 + e ^{10 - t} ) ^{2}} \frac{d}{d t} (1 + e^{10 - t})

= - \frac{1}{( 1 + e ^{10 - t} ) ^{2}} \frac{d}{d t} (1 + e^{10 - t})

\frac{d}{d t} (1 + e^{10 - t}) = - e^{- t + 10}

= 20 (- \frac{1}{( 1 + e ^{10 - t} ) ^{2}} (- e^{- t + 10}))

Simplify

20 (- \frac{1}{( 1 + e ^{10 - t} ) ^{2}} (- e^{- t + 10})) : \frac{20 e ^{- t + 10}}{( 1 + e ^{10 - t} ) ^{2}}

= \frac{20 e ^{- t + 10}}{( 1 + e ^{10 - t} ) ^{2}}

y^{^{'}} = 8 x y

\frac{\partial}{\partial x} (x^{4} + y^{2} - 4)

y^{^{''}} - y^{^{'}} - 2 y = e^{2 t}

\frac{d}{d t} (\frac{1}{t})

\frac{d}{d x} (\frac{1}{( x ^{2} + x + 1 ) ^{3}})

What is the derivative of 20(1+e^{10-t})^{-1} ?
The derivative of 20(1+e^{10-t})^{-1} is (20e^{-t+10})/((1+e^{10-t))^2}
What is the first derivative of 20(1+e^{10-t})^{-1} ?
The first derivative of 20(1+e^{10-t})^{-1} is (20e^{-t+10})/((1+e^{10-t))^2}