What is the derivative of y=80(1-e^{-0.041x}) ?

The derivative of y=80(1-e^{-0.041x}) is 3.28e^{-0.041x}

What is the first derivative of y=80(1-e^{-0.041x}) ?

The first derivative of y=80(1-e^{-0.041x}) is 3.28e^{-0.041x}

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derivative of

y = 80 (1 - e^{- 0.041 x})

3.28 e^{- 0.041 x}

Solution steps

\frac{d}{d x} (80 (1 - e^{- 0.041 x}))

Take the constant out:

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

= 80 \frac{d}{d x} (1 - e^{- 0.041 x})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= 80 (\frac{d}{d x} (1) - \frac{d}{d x} (e^{- 0.041 x}))

\frac{d}{d x} (1) = 0

\frac{d}{d x} (e^{- 0.041 x}) = - 0.041 e^{- 0.041 x}

= 80 (0 - (- 0.041 e^{- 0.041 x}))

Simplify

= 3.28 e^{- 0.041 x}

\frac{\partial}{\partial x} (y (x^{2} + y)^{6})

x \to - 1 lim (x + 5)

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

\frac{d}{d x} (\frac{2}{5 x - 1})

\frac{4}{s}

What is the derivative of y=80(1-e^{-0.041x}) ?
The derivative of y=80(1-e^{-0.041x}) is 3.28e^{-0.041x}
What is the first derivative of y=80(1-e^{-0.041x}) ?
The first derivative of y=80(1-e^{-0.041x}) is 3.28e^{-0.041x}