Turunan dari $$$\ln\left(x + \sqrt{x^{2} + 1}\right)$$$

Temukan $$$\frac{d}{dx} \left(\ln\left(x + \sqrt{x^{2} + 1}\right)\right)$$$.

Fungsi $$$\ln\left(x + \sqrt{x^{2} + 1}\right)$$$ merupakan komposisi $$$f{\left(g{\left(x \right)} \right)}$$$ dari dua fungsi $$$f{\left(u \right)} = \ln\left(u\right)$$$ dan $$$g{\left(x \right)} = x + \sqrt{x^{2} + 1}$$$.

Terapkan aturan rantai $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(x + \sqrt{x^{2} + 1}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x + \sqrt{x^{2} + 1}\right)\right)}$$

Turunan dari logaritma natural adalah $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x + \sqrt{x^{2} + 1}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x + \sqrt{x^{2} + 1}\right)$$

Kembalikan ke variabel semula:

$$\frac{\frac{d}{dx} \left(x + \sqrt{x^{2} + 1}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(x + \sqrt{x^{2} + 1}\right)}{{\color{red}\left(x + \sqrt{x^{2} + 1}\right)}}$$

Turunan dari jumlah/selisih adalah jumlah/selisih dari turunan:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(x + \sqrt{x^{2} + 1}\right)\right)}}{x + \sqrt{x^{2} + 1}} = \frac{{\color{red}\left(\frac{d}{dx} \left(x\right) + \frac{d}{dx} \left(\sqrt{x^{2} + 1}\right)\right)}}{x + \sqrt{x^{2} + 1}}$$

Terapkan aturan pangkat $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ dengan $$$n = 1$$$, dengan kata lain, $$$\frac{d}{dx} \left(x\right) = 1$$$:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(\sqrt{x^{2} + 1}\right)}{x + \sqrt{x^{2} + 1}} = \frac{{\color{red}\left(1\right)} + \frac{d}{dx} \left(\sqrt{x^{2} + 1}\right)}{x + \sqrt{x^{2} + 1}}$$

Fungsi $$$\sqrt{x^{2} + 1}$$$ merupakan komposisi $$$f{\left(g{\left(x \right)} \right)}$$$ dari dua fungsi $$$f{\left(u \right)} = \sqrt{u}$$$ dan $$$g{\left(x \right)} = x^{2} + 1$$$.

Terapkan aturan rantai $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(\sqrt{x^{2} + 1}\right)\right)} + 1}{x + \sqrt{x^{2} + 1}} = \frac{{\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right) \frac{d}{dx} \left(x^{2} + 1\right)\right)} + 1}{x + \sqrt{x^{2} + 1}}$$

Terapkan aturan pangkat $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$ dengan $$$n = \frac{1}{2}$$$:

$$\frac{{\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right)\right)} \frac{d}{dx} \left(x^{2} + 1\right) + 1}{x + \sqrt{x^{2} + 1}} = \frac{{\color{red}\left(\frac{1}{2 \sqrt{u}}\right)} \frac{d}{dx} \left(x^{2} + 1\right) + 1}{x + \sqrt{x^{2} + 1}}$$

Kembalikan ke variabel semula:

$$\frac{1 + \frac{\frac{d}{dx} \left(x^{2} + 1\right)}{2 \sqrt{{\color{red}\left(u\right)}}}}{x + \sqrt{x^{2} + 1}} = \frac{1 + \frac{\frac{d}{dx} \left(x^{2} + 1\right)}{2 \sqrt{{\color{red}\left(x^{2} + 1\right)}}}}{x + \sqrt{x^{2} + 1}}$$

Turunan dari jumlah/selisih adalah jumlah/selisih dari turunan:

$$\frac{1 + \frac{{\color{red}\left(\frac{d}{dx} \left(x^{2} + 1\right)\right)}}{2 \sqrt{x^{2} + 1}}}{x + \sqrt{x^{2} + 1}} = \frac{1 + \frac{{\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(1\right)\right)}}{2 \sqrt{x^{2} + 1}}}{x + \sqrt{x^{2} + 1}}$$

Turunan dari suatu konstanta adalah $$$0$$$:

$$\frac{1 + \frac{{\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x^{2}\right)}{2 \sqrt{x^{2} + 1}}}{x + \sqrt{x^{2} + 1}} = \frac{1 + \frac{{\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{2}\right)}{2 \sqrt{x^{2} + 1}}}{x + \sqrt{x^{2} + 1}}$$

Terapkan aturan pangkat $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ dengan $$$n = 2$$$:

$$\frac{1 + \frac{{\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)}}{2 \sqrt{x^{2} + 1}}}{x + \sqrt{x^{2} + 1}} = \frac{1 + \frac{{\color{red}\left(2 x\right)}}{2 \sqrt{x^{2} + 1}}}{x + \sqrt{x^{2} + 1}}$$

Sederhanakan:

$$\frac{\frac{x}{\sqrt{x^{2} + 1}} + 1}{x + \sqrt{x^{2} + 1}} = \frac{1}{\sqrt{x^{2} + 1}}$$

Dengan demikian, $$$\frac{d}{dx} \left(\ln\left(x + \sqrt{x^{2} + 1}\right)\right) = \frac{1}{\sqrt{x^{2} + 1}}$$$.