Whakaoti i te (y^4)^20=

$$( y ^ { 4 } ) ^ { 20 } =$$

Aromātai

$$y^{80}$$

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

Whakamahia ngā ture taupū hei whakarūnā i te kīanga.

$$\left(y^{4}\right)^{20}$$

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

$$y^{4\times 20}$$

Whakareatia 4 ki te 20.

$$y^{80}$$

Kimi Pārōnaki e ai ki y

$$80y^{79}$$

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).

$$20\left(y^{4}\right)^{20-1}\frac{\mathrm{d}}{\mathrm{d}y}(y^{4})$$

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}.

$$20\left(y^{4}\right)^{19}\times 4y^{4-1}$$

Whakarūnātia.

$$80y^{3}\left(y^{4}\right)^{19}$$