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\frac{\mathrm{d}}{\mathrm{d}\beta }(\sin(\beta ))=\left(\lim_{h\to 0}\frac{\sin(\beta +h)-\sin(\beta )}{h}\right)
f\left(x\right) funksiyasi uchun, hosilasi \frac{f\left(x+h\right)-f\left(x\right)}{h} cheklovidir, chunki ana shu cheklov mavjud bo'lsa, h 0'ga o'tadi.
\lim_{h\to 0}\frac{\sin(h+\beta )-\sin(\beta )}{h}
Sinus uchun yig'indi formulasidan foydalanish.
\lim_{h\to 0}\frac{\sin(\beta )\left(\cos(h)-1\right)+\cos(\beta )\sin(h)}{h}
\sin(\beta ) omili.
\left(\lim_{h\to 0}\sin(\beta )\right)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\left(\lim_{h\to 0}\cos(\beta )\right)\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)
Chegarani qayta yozish.
\sin(\beta )\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(\beta )\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)
Limitlar h dan 0 sifatida hisoblanganda \beta ni konstanta sifatida foydalanish.
\sin(\beta )\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(\beta )
\lim_{\beta \to 0}\frac{\sin(\beta )}{\beta } chegarasi 1 dir.
\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)
\lim_{h\to 0}\frac{\cos(h)-1}{h} chegarasini baholash uchun, avval surat va maxrajni \cos(h)+1 ga ko'paytiring.
\lim_{h\to 0}\frac{\left(\cos(h)\right)^{2}-1}{h\left(\cos(h)+1\right)}
\cos(h)+1 ni \cos(h)-1 marotabaga ko'paytirish.
\lim_{h\to 0}-\frac{\left(\sin(h)\right)^{2}}{h\left(\cos(h)+1\right)}
Pifagor ayniyatidan foydalanish.
\left(\lim_{h\to 0}-\frac{\sin(h)}{h}\right)\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)
Chegarani qayta yozish.
-\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)
\lim_{\beta \to 0}\frac{\sin(\beta )}{\beta } chegarasi 1 dir.
\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)=0
\frac{\sin(h)}{\cos(h)+1} 0 da davomiy sifatida foydalanish.
\cos(\beta )
0 qiymatini \sin(\beta )\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(\beta ) ifodasiga almashtirish.