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\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))
2 va 2 ni qisqartiring.
\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x))=\left(\lim_{h\to 0}\frac{\cos(x+h)-\cos(x)}{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{\cos(x+h)-\cos(x)}{h}
Kosinus uchun yig'indi formulasidan foydalanish.
\lim_{h\to 0}\frac{\cos(x)\left(\cos(h)-1\right)-\sin(x)\sin(h)}{h}
\cos(x) omili.
\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)
Chegarani qayta yozish.
\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)
Limitlar h dan 0 sifatida hisoblanganda x ni konstanta sifatida foydalanish.
\cos(x)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(x)
\lim_{x\to 0}\frac{\sin(x)}{x} 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_{x\to 0}\frac{\sin(x)}{x} 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.
-\sin(x)
0 qiymatini \cos(x)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(x) ifodasiga almashtirish.