What is the derivative of 20+53e^{kt} ?

The derivative of 20+53e^{kt} is 53e^{kt}t

What is the first derivative of 20+53e^{kt} ?

The first derivative of 20+53e^{kt} is 53e^{kt}t

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derivative of

20 + 53 e^{k t}

53 e^{k t} t

Solution steps

\frac{d}{d k} (20 + 53 e^{k t})

Treat

t

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d k} (20) + \frac{d}{d k} (53 e^{k t})

\frac{d}{d k} (20) = 0

\frac{d}{d k} (53 e^{k t}) = 53 e^{k t} t

= 0 + 53 e^{k t} t

Simplify

= 53 e^{k t} t

\frac{3 x - 1}{( x - 1 ) ^{3}}

\int 6 x + 2 d x

(x + 2)^{3} - e^{3 x}

y^{^{''}} + 7 y^{^{'}} + 10 y = - 20 t e^{4 t}

f (x) = \frac{1}{1 + e ^{- x}}

What is the derivative of 20+53e^{kt} ?
The derivative of 20+53e^{kt} is 53e^{kt}t
What is the first derivative of 20+53e^{kt} ?
The first derivative of 20+53e^{kt} is 53e^{kt}t