Whakaoti i te [(x)^3]^4=

$$[ ( x ) ^ { 3 } ] ^ { 4 } =$$

Aromātai

$$x^{12}$$

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

Whakamahia ngā ture taupū hei whakarūnā i te kīanga.

$$\left(x^{3}\right)^{4}$$

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

$$x^{3\times 4}$$

Whakareatia 3 ki te 4.

$$x^{12}$$

Kimi Pārōnaki e ai ki x

$$12x^{11}$$

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).

$$4\left(x^{3}\right)^{4-1}\frac{\mathrm{d}}{\mathrm{d}x}(x^{3})$$

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}.

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

Whakarūnātia.

$$12x^{2}\left(x^{3}\right)^{3}$$