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

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

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

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

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derivative of

ln (t + 1)

\frac{1}{t + 1}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

= \frac{1}{t + 1}

\frac{\partial}{\partial x} (e^{- 3 y} cos (- 3 x))

\int \frac{1}{1 + t ^{2}} d t

\int_{1}^{x} 2 (t - 1) d t

x^{2} y^{^{''}} - 3 x y^{^{'}} + 4 y = x^{\frac{5}{2}}

\frac{\partial}{\partial x} (7 ln (\frac{3 x y}{7 z}))

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