Whakaoti i te (n^2)^4

$$( n ^ { 2 } ) ^ { 4 }$$

Aromātai

$$n^{8}$$

Ngā Upane mō te Hāpai i Tētahi Pū ki Tētahi Atu Pū

Whakamahia ngā ture taupū hei whakarūnā i te kīanga.

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

Hei hiki pū ki tētahi pū anō, me whakarea ngā taupū.

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

Whakareatia 2 ki te 4.

$$n^{8}$$

Kimi Pārōnaki e ai ki n

$$8n^{7}$$

Ngā Upane e Whakamahi ana i te Ture Mekameka

Mēnā ko F te hanganga o ngā pānga e rua e taea ana te pārōnaki f\left(u\right) me u=g\left(x\right), arā, mēnā ko F\left(x\right)=f\left(g\left(x\right)\right), ko te pārōnaki o F te pārōnaki o f e ai ki u whakareatia te pārōnaki o g e ai ki x, arā, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

$$4\left(n^{2}\right)^{4-1}\frac{\mathrm{d}}{\mathrm{d}n}(n^{2})$$

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}.

$$4\left(n^{2}\right)^{3}\times 2n^{2-1}$$

Whakarūnātia.

$$8n^{1}\left(n^{2}\right)^{3}$$

Mō tētahi kupu t, t^{1}=t.

$$8n\left(n^{2}\right)^{3}$$