Turunan dari $$$\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$

Kalkulator akan menentukan turunan dari $$$\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$, dengan langkah-langkah yang ditampilkan.

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Masukan Anda

Temukan $$$\frac{d}{du} \left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right)$$$.

Solusi

Fungsi $$$\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$ merupakan komposisi $$$f{\left(g{\left(u \right)} \right)}$$$ dari dua fungsi $$$f{\left(v \right)} = \tan{\left(v \right)}$$$ dan $$$g{\left(u \right)} = \frac{u}{2} + \frac{\pi}{4}$$$.

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

$${\color{red}\left(\frac{d}{du} \left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right)\right)} = {\color{red}\left(\frac{d}{dv} \left(\tan{\left(v \right)}\right) \frac{d}{du} \left(\frac{u}{2} + \frac{\pi}{4}\right)\right)}$$

Turunan fungsi tangen adalah $$$\frac{d}{dv} \left(\tan{\left(v \right)}\right) = \sec^{2}{\left(v \right)}$$$:

$${\color{red}\left(\frac{d}{dv} \left(\tan{\left(v \right)}\right)\right)} \frac{d}{du} \left(\frac{u}{2} + \frac{\pi}{4}\right) = {\color{red}\left(\sec^{2}{\left(v \right)}\right)} \frac{d}{du} \left(\frac{u}{2} + \frac{\pi}{4}\right)$$

Kembalikan ke variabel semula:

$$\sec^{2}{\left({\color{red}\left(v\right)} \right)} \frac{d}{du} \left(\frac{u}{2} + \frac{\pi}{4}\right) = \sec^{2}{\left({\color{red}\left(\frac{u}{2} + \frac{\pi}{4}\right)} \right)} \frac{d}{du} \left(\frac{u}{2} + \frac{\pi}{4}\right)$$

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

$$\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{du} \left(\frac{u}{2} + \frac{\pi}{4}\right)\right)} = \sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{du} \left(\frac{u}{2}\right) + \frac{d}{du} \left(\frac{\pi}{4}\right)\right)}$$

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

$$\left({\color{red}\left(\frac{d}{du} \left(\frac{\pi}{4}\right)\right)} + \frac{d}{du} \left(\frac{u}{2}\right)\right) \sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} = \left({\color{red}\left(0\right)} + \frac{d}{du} \left(\frac{u}{2}\right)\right) \sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$

Terapkan aturan kelipatan konstanta $$$\frac{d}{du} \left(c f{\left(u \right)}\right) = c \frac{d}{du} \left(f{\left(u \right)}\right)$$$ dengan $$$c = \frac{1}{2}$$$ dan $$$f{\left(u \right)} = u$$$:

$$\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{du} \left(\frac{u}{2}\right)\right)} = \sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{\frac{d}{du} \left(u\right)}{2}\right)}$$

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

$$\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} {\color{red}\left(\frac{d}{du} \left(u\right)\right)}}{2} = \frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} {\color{red}\left(1\right)}}{2}$$

Sederhanakan:

$$\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2} = \frac{1}{1 - \sin{\left(u \right)}}$$

Dengan demikian, $$$\frac{d}{du} \left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right) = \frac{1}{1 - \sin{\left(u \right)}}$$$.

Jawaban

$$$\frac{d}{du} \left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right) = \frac{1}{1 - \sin{\left(u \right)}}$$$A


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