Whakaoti i te (n-2sqrt{2})(n+2sqrt{2})

$$( n - 2 \sqrt { 2 } ) ( n + 2 \sqrt { 2 } )$$

Aromātai

$$n^{2}-8$$

Ngā Upane Otinga

Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$n^{2}-\left(2\sqrt{2}\right)^{2}$$

Whakarohaina te \left(2\sqrt{2}\right)^{2}.

$$n^{2}-2^{2}\left(\sqrt{2}\right)^{2}$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$n^{2}-4\left(\sqrt{2}\right)^{2}$$

Ko te pūrua o \sqrt{2} ko 2.

$$n^{2}-4\times 2$$

Whakareatia te 4 ki te 2, ka 8.

$$n^{2}-8$$

Kimi Pārōnaki e ai ki n

$$2n$$

Ngā Upane e Whakamahi ana i te Ture Pārōnaki mō te Tapeke

Whakaarohia te \left(n-2\sqrt{2}\right)\left(n+2\sqrt{2}\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}n}(n^{2}-\left(2\sqrt{2}\right)^{2})$$

Whakarohaina te \left(2\sqrt{2}\right)^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}n}(n^{2}-2^{2}\left(\sqrt{2}\right)^{2})$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$\frac{\mathrm{d}}{\mathrm{d}n}(n^{2}-4\left(\sqrt{2}\right)^{2})$$

Ko te pūrua o \sqrt{2} ko 2.

$$\frac{\mathrm{d}}{\mathrm{d}n}(n^{2}-4\times 2)$$

Whakareatia te 4 ki te 2, ka 8.

$$\frac{\mathrm{d}}{\mathrm{d}n}(n^{2}-8)$$

Ko te pārōnaki o tētahi pūrau ko te tapeke o ngā pārōnaki o ōna kīanga tau. Ko te pārōnaki o tētahi kīanga tau pūmau ko 0. Ko te pārōnaki o te ax^{n} ko te nax^{n-1}.

$$2n^{2-1}$$

Tango 1 mai i 2.

$$2n^{1}$$

Mō tētahi kupu t, t^{1}=t.

$$2n$$