What is the derivative of ln(t+1)-t ?

The derivative of ln(t+1)-t is 1/(t+1)-1

What is the first derivative of ln(t+1)-t ?

The first derivative of ln(t+1)-t is 1/(t+1)-1

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derivative of

ln (t + 1) - t

\frac{1}{t + 1} - 1

Solution steps

\frac{d}{d t} (ln (t + 1) - t)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (ln (t + 1)) - \frac{d t}{d t}

\frac{d}{d t} (ln (t + 1)) = \frac{1}{t + 1}

\frac{d t}{d t} = 1

= \frac{1}{t + 1} - 1

\frac{d}{d n} (\frac{( n + 1 ) ^{n}}{n !})

\frac{1}{ln ( 1 + 3 x )}

lo g_{10} (tan (\frac{a}{2}))

x^{2} - 3 x + 1

x = 2 t^{2}

What is the derivative of ln(t+1)-t ?
The derivative of ln(t+1)-t is 1/(t+1)-1
What is the first derivative of ln(t+1)-t ?
The first derivative of ln(t+1)-t is 1/(t+1)-1