# Derivative of $\ln\left(x + 1\right)$

The calculator will find the derivative of $\ln\left(x + 1\right)$, with steps shown.

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Find $\frac{d}{dx} \left(\ln\left(x + 1\right)\right)$.

### Solution

The function $\ln\left(x + 1\right)$ is the composition $f{\left(g{\left(x \right)} \right)}$ of two functions $f{\left(u \right)} = \ln\left(u\right)$ and $g{\left(x \right)} = x + 1$.

Apply the chain rule $\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$:

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(x + 1\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x + 1\right)\right)}$$

The derivative of the natural logarithm is $\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$:

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x + 1\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x + 1\right)$$

$$\frac{\frac{d}{dx} \left(x + 1\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(x + 1\right)}{{\color{red}\left(x + 1\right)}}$$

The derivative of a sum/difference is the sum/difference of derivatives:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(x + 1\right)\right)}}{x + 1} = \frac{{\color{red}\left(\frac{d}{dx} \left(x\right) + \frac{d}{dx} \left(1\right)\right)}}{x + 1}$$

The derivative of a constant is $0$:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x\right)}{x + 1} = \frac{{\color{red}\left(0\right)} + \frac{d}{dx} \left(x\right)}{x + 1}$$

Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 1$, in other words, $\frac{d}{dx} \left(x\right) = 1$:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{x + 1} = \frac{{\color{red}\left(1\right)}}{x + 1}$$

Thus, $\frac{d}{dx} \left(\ln\left(x + 1\right)\right) = \frac{1}{x + 1}$.

$\frac{d}{dx} \left(\ln\left(x + 1\right)\right) = \frac{1}{x + 1}$A