Whakaoti i te (u^-4)^3

$$( u ^ { - 4 } ) ^ { 3 }$$

Aromātai

$$\frac{1}{u^{12}}$$

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

Whakamahia ngā ture taupū hei whakarūnā i te kīanga.

$$\left(u^{-4}\right)^{3}$$

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

$$u^{-4\times 3}$$

Whakareatia -4 ki te 3.

$$\frac{1}{u^{12}}$$

Kimi Pārōnaki e ai ki u

$$-\frac{12}{u^{13}}$$

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).

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

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}.

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

Whakarūnātia.

$$-12u^{-5}\left(u^{-4}\right)^{2}$$