Whakaoti i te d/dxsinx=

$$\frac { d } { d x } \sin x =$$

Aromātai

$$\cos(x)$$

Ngā Upane e Whakamahi ana i te Tautuhinga o te Pārōnaki

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.

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

Whakamahia te Tikanga Tātai Tapeke mō te Aho.

$$\lim_{h\to 0}\frac{\sin(x+h)-\sin(x)}{h}$$

Tauwehea te \sin(x).

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

Tuhia anō te tepe.

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

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

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

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

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

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

$$\cos(x)$$

Kimi Pārōnaki e ai ki x

$$-\sin(x)$$