|
Derivative of $$$\ln\left(x + 1\right)$$$
Related calculators: Logarithmic Differentiation Calculator, Implicit Differentiation Calculator with Steps
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