Whakaoti i te cosθ/1

$$\frac { \cos \theta } { 1 }$$

Kimi Pārōnaki e ai ki θ

$$-\sin(\theta )$$

Ngā Upane e Whakamahi ana i te Tautuhinga o te Pārōnaki

Ka whakawehea he tau ki te tahi, hua ai ko ia anō.

$$\frac{\mathrm{d}}{\mathrm{d}\theta }(\cos(\theta ))$$

Mō tētahi pānga f\left(x\right), ko te pārōnaki te tepe o \frac{f\left(x+h\right)-f\left(x\right)}{h} ina haere h ki 0, mēnā kei reira taua tepe.

$$\frac{\mathrm{d}}{\mathrm{d}\theta }(\cos(\theta ))=\left(\lim_{h\to 0}\frac{\cos(\theta +h)-\cos(\theta )}{h}\right)$$

Whakamahia te Tikanga Tātai Tapeke mō te Whenu.

$$\lim_{h\to 0}\frac{\cos(h+\theta )-\cos(\theta )}{h}$$

Tauwehea te \cos(\theta ).

$$\lim_{h\to 0}\frac{\cos(\theta )\left(\cos(h)-1\right)-\sin(\theta )\sin(h)}{h}$$

Tuhia anō te tepe.

$$\left(\lim_{h\to 0}\cos(\theta )\right)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\left(\lim_{h\to 0}\sin(\theta )\right)\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)$$

Whakamahia te meka ko \theta he pūmau ina tātai tepe i te wā ka haere h ki te 0.

$$\cos(\theta )\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(\theta )\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)$$

Ko te tepe \lim_{\theta \to 0}\frac{\sin(\theta )}{\theta } he 1.

$$\cos(\theta )\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(\theta )$$

Hei arotake i te tepe \lim_{h\to 0}\frac{\cos(h)-1}{h}, tuatahi me whakarea te taurunga me te tauraro ki te \cos(h)+1.

$$\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)=\left(\lim_{h\to 0}\frac{\left(\cos(h)-1\right)\left(\cos(h)+1\right)}{h\left(\cos(h)+1\right)}\right)$$

Whakareatia \cos(h)+1 ki te \cos(h)-1.

$$\lim_{h\to 0}\frac{\left(\cos(h)\right)^{2}-1}{h\left(\cos(h)+1\right)}$$

Whakamahia te Tuakiri Pythagorean.

$$\lim_{h\to 0}-\frac{\left(\sin(h)\right)^{2}}{h\left(\cos(h)+1\right)}$$

Tuhia anō te tepe.

$$\left(\lim_{h\to 0}-\frac{\sin(h)}{h}\right)\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)$$

Ko te tepe \lim_{\theta \to 0}\frac{\sin(\theta )}{\theta } he 1.

$$-\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)$$

Whakamahia te meka he motukore a \frac{\sin(h)}{\cos(h)+1} i 0.

$$\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)=0$$

Whakakapihia te uara 0 ki roto i te kīanga \cos(\theta )\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(\theta ).

$$-\sin(\theta )$$

Aromātai

$$\cos(\theta )$$