What is the derivative of e^{ln(1+1/k)} ?

The derivative of e^{ln(1+1/k)} is -1/(k^2)

What is the first derivative of e^{ln(1+1/k)} ?

The first derivative of e^{ln(1+1/k)} is -1/(k^2)

Popular >

derivative of

e^{l n (1 + \frac{1}{k})}

- \frac{1}{k ^{2}}

Solution steps

\frac{d}{d k} (e^{l n (1 + \frac{1}{k})})

Simplify

e^{l n (1 + \frac{1}{k})} : 1 + \frac{1}{k}

= \frac{d}{d k} (1 + \frac{1}{k})

Apply the Sum/Difference Rule:

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

= \frac{d}{d k} (1) + \frac{d}{d k} (\frac{1}{k})

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

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

= 0 - \frac{1}{k ^{2}}

Simplify

= - \frac{1}{k ^{2}}

\int_{0}^{2 π} t^{2} sin (2 t) d t

f (x) = \frac{32}{x ^{3}}

x + 2, 1, 3

\frac{d}{d x} (x ln (x) cos (x))

\int \frac{1}{3 x} + cos (2 x) - 5 x^{2} d x

What is the derivative of e^{ln(1+1/k)} ?
The derivative of e^{ln(1+1/k)} is -1/(k^2)
What is the first derivative of e^{ln(1+1/k)} ?
The first derivative of e^{ln(1+1/k)} is -1/(k^2)