ແກ້ sin(x2)

$$\sin ( x2 )$$

ບອກຄວາມແຕກຕ່າງ w.r.t. x_2

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

ຂັ້ນຕອນການໃຊ້ຄຳນິຍາມຂອງອະນຸພັນ

ສຳລັບຟັງຊັນ f\left(x\right), ອະນຸພັນແມ່ນຂໍ້ຈຳກັດຂອງ \frac{f\left(x+h\right)-f\left(x\right)}{h} ເມື່ອ h ໄປເປັນ 0, ຫາກມີຂໍ້ຈຳກັດນັ້ນ.

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

ໃຊ້ສູດບວກຮວມສຳລັບຊິນ.

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

ຕົວປະກອບຈາກ \sin(x_{2}).

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

ຂຽນຂໍ້ຈຳກັດຄືນໃໝ່.

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

ໃຊ້ຂໍ້ເທັດຈິງທີ່ x_{2} ແມ່ນຄົງທີ່ເມື່ອການຄຳນວນຈຳກັດເປັນ h ໄປທີ່ 0.

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

ຂໍ້ຈຳກັດ \lim_{x_{2}\to 0}\frac{\sin(x_{2})}{x_{2}} ແມ່ນ 1.

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

ເພື່ອປະເມີນຂໍ້ຈຳກັດ \lim_{h\to 0}\frac{\cos(h)-1}{h}, ທຳອິດໃຫ້ຄູນຕົວເສດ ແລະ ຕົວຫານດ້ວຍ \cos(h)+1.

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

ຄູນ \cos(h)+1 ໃຫ້ກັບ \cos(h)-1.

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

ໃຊ້ການລະບຸຕົວຕົນທິດສະດີປີຕາກໍຣັສ.

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

ຂຽນຂໍ້ຈຳກັດຄືນໃໝ່.

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

ຂໍ້ຈຳກັດ \lim_{x_{2}\to 0}\frac{\sin(x_{2})}{x_{2}} ແມ່ນ 1.

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

ໃຊ້ຂໍ້ເທັດຈິງທີ່ \frac{\sin(h)}{\cos(h)+1} ແມ່ນຕໍ່ເນື່ອງທີ່ 0.

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

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

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

ປະເມີນ

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