Kimi Pārōnaki e ai ki x
-\sin(x)
Aromātai
\cos(x)
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Kua tāruatia ki te papatopenga
\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))=\left(\lim_{h\to 0}\frac{\cos(x+h)-\cos(x)}{h}\right)
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.
\lim_{h\to 0}\frac{\cos(x+h)-\cos(x)}{h}
Whakamahia te Tikanga Tātai Tapeke mō te Whenu.
\lim_{h\to 0}\frac{\cos(x)\left(\cos(h)-1\right)-\sin(x)\sin(h)}{h}
Tauwehea te \cos(x).
\left(\lim_{h\to 0}\cos(x)\right)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\left(\lim_{h\to 0}\sin(x)\right)\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)
Tuhia anō te tepe.
\cos(x)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(x)\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.
\cos(x)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(x)
Ko te tepe \lim_{x\to 0}\frac{\sin(x)}{x} he 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)
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.
\lim_{h\to 0}\frac{\left(\cos(h)\right)^{2}-1}{h\left(\cos(h)+1\right)}
Whakareatia \cos(h)+1 ki te \cos(h)-1.
\lim_{h\to 0}-\frac{\left(\sin(h)\right)^{2}}{h\left(\cos(h)+1\right)}
Whakamahia te Tuakiri Pythagorean.
\left(\lim_{h\to 0}-\frac{\sin(h)}{h}\right)\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)
Tuhia anō te tepe.
-\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)=0
Whakamahia te meka he motukore a \frac{\sin(h)}{\cos(h)+1} i 0.
-\sin(x)
Whakakapihia te uara 0 ki roto i te kīanga \cos(x)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(x).