What is the derivative of f(x)=30+10e^{-0.05x} ?

The derivative of f(x)=30+10e^{-0.05x} is -0.5e^{-0.05x}

What is the first derivative of f(x)=30+10e^{-0.05x} ?

The first derivative of f(x)=30+10e^{-0.05x} is -0.5e^{-0.05x}

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derivative of

f (x) = 30 + 10 e^{- 0.05 x}

- 0.5 e^{- 0.05 x}

Solution steps

\frac{d}{d x} (30 + 10 e^{- 0.05 x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (30) + \frac{d}{d x} (10 e^{- 0.05 x})

\frac{d}{d x} (30) = 0

\frac{d}{d x} (10 e^{- 0.05 x}) = - 0.5 e^{- 0.05 x}

= 0 - 0.5 e^{- 0.05 x}

Simplify

= - 0.5 e^{- 0.05 x}

cos^{2} (t)

\frac{\partial}{\partial x} (7 e^{x y} \cdot cos (x yz))

x \to 0 lim (\frac{f ( x )}{x})

\frac{d y}{d x} = y (1000 - y)

y = 5 e^{- x}, (0, 5)

What is the derivative of f(x)=30+10e^{-0.05x} ?
The derivative of f(x)=30+10e^{-0.05x} is -0.5e^{-0.05x}
What is the first derivative of f(x)=30+10e^{-0.05x} ?
The first derivative of f(x)=30+10e^{-0.05x} is -0.5e^{-0.05x}