Whakaoti i te (z^2)^2/(z^6)^8=

$$\frac { ( z ^ { 2 } ) ^ { 2 } } { ( z ^ { 6 } ) ^ { 8 } } =$$

Aromātai

$$\frac{1}{z^{44}}$$

Ngā Upane Otinga

Hei hiki pū ki tētahi pū anō, me whakarea ngā taupū. Me whakarea te 2 me te 2 kia riro ai te 4.

$$\frac{z^{4}}{\left(z^{6}\right)^{8}}$$

Hei hiki pū ki tētahi pū anō, me whakarea ngā taupū. Me whakarea te 6 me te 8 kia riro ai te 48.

$$\frac{z^{4}}{z^{48}}$$

Tuhia anō te z^{48} hei z^{4}z^{44}. Me whakakore tahi te z^{4} i te taurunga me te tauraro.

$$\frac{1}{z^{44}}$$

Kimi Pārōnaki e ai ki z

$$-\frac{44}{z^{45}}$$

Ngā Upane e Whakamahi ana i te Ture Mekameka

Hei hiki pū ki tētahi pū anō, me whakarea ngā taupū. Me whakarea te 2 me te 2 kia riro ai te 4.

$$\frac{\mathrm{d}}{\mathrm{d}z}(\frac{z^{4}}{\left(z^{6}\right)^{8}})$$

Hei hiki pū ki tētahi pū anō, me whakarea ngā taupū. Me whakarea te 6 me te 8 kia riro ai te 48.

$$\frac{\mathrm{d}}{\mathrm{d}z}(\frac{z^{4}}{z^{48}})$$

Tuhia anō te z^{48} hei z^{4}z^{44}. Me whakakore tahi te z^{4} i te taurunga me te tauraro.

$$\frac{\mathrm{d}}{\mathrm{d}z}(\frac{1}{z^{44}})$$

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).

$$-\left(z^{44}\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}z}(z^{44})$$

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}.

$$-\left(z^{44}\right)^{-2}\times 44z^{44-1}$$

Whakarūnātia.

$$-44z^{43}\left(z^{44}\right)^{-2}$$