Datrys sin(x2)

$$\sin ( x2 )$$

Gwahaniaethu w.r.t. x_2

$$\cos(x_{2})$$

Camau gan Ddefnyddio Diffiniad o Ddeilliad

Ar gyfer y ffwythiant f\left(x\right), y deilliad yw'r terfyn \frac{f\left(x+h\right)-f\left(x\right)}{h} gan fod h yn mynd i 0, os yw’r terfyn hwnnw yn bodoli.

$$\frac{\mathrm{d}}{\mathrm{d}x_{2}}(\sin(x_{2}))=\left(\lim_{h\to 0}\frac{\sin(x_{2}+h)-\sin(x_{2})}{h}\right)$$

Defnyddio’r Fformiwla Swm ar gyfer Sin.

$$\lim_{h\to 0}\frac{\sin(x_{2}+h)-\sin(x_{2})}{h}$$

Ffactora allan \sin(x_{2}).

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

Ailysgrifennu’r terfyn.

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

Defnyddio'r ffaith bod x_{2} yn gyson pan fydd terfynau cyfrifiadura fel h yn mynd i 0.

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

Y terfyn \lim_{x_{2}\to 0}\frac{\sin(x_{2})}{x_{2}} yw 1.

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

I enrhifo’r terfyn \lim_{h\to 0}\frac{\cos(h)-1}{h}, yn gyntaf lluoswch y rhifiadur a'r enwadur â \cos(h)+1.

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

Lluoswch \cos(h)+1 â \cos(h)-1.

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

Defnyddio'r Hunaniaeth Pythagoreaidd.

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

Ailysgrifennu’r terfyn.

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

Y terfyn \lim_{x_{2}\to 0}\frac{\sin(x_{2})}{x_{2}} yw 1.

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

Defnyddio'r ffaith fod \frac{\sin(h)}{\cos(h)+1} yn barhaus yn 0.

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

Amnewid y gwerth 0 i mewn i’r mynegiad \sin(x_{2})\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(x_{2}).

$$\cos(x_{2})$$

Enrhifo

$$\sin(x_{2})$$