tan(x)の微分 2(微分の定義からの場合)

$\tan \left(x\right)$の微分 2

(sin, cosの微分を既知とした場合) $$\begin{array}{r}\\ y& =& \tan \left(x\right)\\ \frac{\mathrm{d}}{\mathrm{d}x}\tan \left(x\right)& =& \underset{h\to 0}{lim}\frac{\tan (x+h)-\tan \left(x\right)}{h}\\ & =& \underset{h\to 0}{lim}\frac{1}{h}\{\frac{\tan \left(x\right)+\tan \left(h\right)}{1-\tan \left(x\right)\tan \left(h\right)}-\tan \left(x\right)\}\\ & =& \underset{h\to 0}{lim}\frac{1}{h}\frac{\tan \left(x\right)+\tan \left(h\right)-\tan \left(x\right)(1-\tan \left(x\right)\tan \left(h\right))}{1-\tan \left(x\right)\tan \left(h\right)}\\ & =& \underset{h\to 0}{lim}\frac{1}{h}\frac{\tan \left(x\right)+\tan \left(h\right)-\tan \left(x\right)+{\tan }^2\left(x\right)\tan \left(h\right)}{1-\tan \left(x\right)\tan \left(h\right)}\\ & =& \underset{h\to 0}{lim}\frac{1}{h}\frac{\tan \left(h\right)+{\tan }^2\left(x\right)\tan \left(h\right)}{1-\tan \left(x\right)\tan \left(h\right)}\\ & =& \underset{h\to 0}{lim}\frac{1}{h}\frac{\tan \left(h\right)(1+{\tan }^2\left(x\right))}{1-\tan \left(x\right)\tan \left(h\right)}\\ & =& \underset{h\to 0}{lim}\frac{\tan \left(h\right)}{h}\frac{1+{\tan }^2\left(x\right)}{1-\tan \left(x\right)\tan \left(h\right)}\\ & =& 1\cdot \frac{1+{\tan }^2\left(x\right)}{1-\tan \left(x\right)\cdot 0}\\ & & \dots \underset{h\to 0}{lim}\frac{\tan \left(h\right)}{h}=\underset{h\to 0}{lim}\frac{1}{h}\frac{\sin \left(h\right)}{\cos \left(h\right)}=\underset{h\to 0}{lim}\frac{\sin \left(h\right)}{h}\frac{1}{\cos \left(h\right)}=1\cdot 1=1\\ & & \dots \underset{h\to 0}{lim}\frac{\sin \left(h\right)}{h}=1\\ & & \dots \underset{h\to 0}{lim}\cos \left(h\right)=1\\ & =& 1+{\tan }^2\left(x\right)\end{array}$$ $$\begin{array}{rcl}1+{\tan }^2\left(x\right)& =& 1+{\left\{\frac{\sin \left(x\right)}{\cos \left(x\right)}\right\}}^2\\ & =& \frac{{\cos }^2\left(x\right)+{\sin }^2\left(x\right)}{{\cos }^2\left(x\right)}\\ & =& \frac{1}{{\cos }^2\left(x\right)}\cdots {\cos }^2\left(x\right)+{\sin }^2\left(x\right)=1\\ & =& {\sec }^2\left(x\right)\end{array}$$