Derivative of ln(x + 1)

The calculator will find the derivative of

\ln\left(x + 1\right)

, with steps shown.

Your Input

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)

Return to the old variable:

\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}

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

The derivative of a constant is $0$ :

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

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

Answer

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

Derivative of ln⁡(x+1)\ln\left(x + 1\right)ln(x+1)

Your Input

Solution

Answer

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