What is the Derivative of ln(x+1)?

The Mathematician — Wed, 02 Nov 2022 13:00:41 +0000

The derivative of

\ln(x+1)

\frac{1}{x + 1}

.

Solution. Let

F(x) = \ln(x+1)

f(u) = \ln(u)

and

g(x) = x + 1

. Then we will use the chain rule:

\begin{align*} F'(x) = f'(g(x))g'(x). \end{align*}

We have seen in here that

\frac{d}{dx} \ln(x) = \frac{1}{x}

. So we get:

\begin{align*} f'(u) = \frac{1}{u} \quad \text{and} \quad g'(x) = 1. \end{align*}

Therefore, we get:

\begin{align*} F'(x) &= f'(g(x))g'(x) \\ &= \frac{1}{x + 1} \cdot 1 \\ &= \frac{1}{x + 1}. \end{align*}

So, the derivative of

\ln(x+1)

\frac{1}{x + 1}

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