cosn eching

$$\cos n$$

n ga nisbatan hosilani topish

$$-\sin(n)$$

Derivativ ifodasidan foydalanish qadamlari

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.

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

Kosinus uchun yig'indi formulasidan foydalanish.

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

\cos(n) omili.

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

Chegarani qayta yozish.

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

Limitlar h dan 0 sifatida hisoblanganda n ni konstanta sifatida foydalanish.

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

\lim_{n\to 0}\frac{\sin(n)}{n} chegarasi 1 dir.

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

\lim_{h\to 0}\frac{\cos(h)-1}{h} chegarasini baholash uchun, avval surat va maxrajni \cos(h)+1 ga ko'paytiring.

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

\cos(h)+1 ni \cos(h)-1 marotabaga ko'paytirish.

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

Pifagor ayniyatidan foydalanish.

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

Chegarani qayta yozish.

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

\lim_{n\to 0}\frac{\sin(n)}{n} chegarasi 1 dir.

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

\frac{\sin(h)}{\cos(h)+1} 0 da davomiy sifatida foydalanish.

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

0 qiymatini \cos(n)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(n) ifodasiga almashtirish.

$$-\sin(n)$$

Baholash

$$\cos(n)$$