Whakaoti i te (64x^9)^1/3

$$( 64 x ^ { 9 } ) ^ { 1 / 3 }$$

Aromātai

$$4x^{3}$$

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

Whakamahia ngā ture taupū hei whakarūnā i te kīanga.

$$\sqrt[3]{64x^{9}}$$

Hei hiki i te hua o ngā tau e rua, neke atu rānei ki tētahi pū, hīkina ia tau ki te pū ka tuhi ko tāna hua.

$$\sqrt[3]{64}\sqrt[3]{x^{9}}$$

Hīkina te 64 ki te pū \frac{1}{3}.

$$4\sqrt[3]{x^{9}}$$

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

$$4x^{9\times \frac{1}{3}}$$

Whakareatia 9 ki te \frac{1}{3}.

$$4x^{3}$$

Kimi Pārōnaki e ai ki x

$$12x^{2}$$

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).

$$\frac{1}{3}\times \left(64x^{9}\right)^{\frac{1}{3}-1}\frac{\mathrm{d}}{\mathrm{d}x}(64x^{9})$$

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}.

$$\frac{1}{3}\times \left(64x^{9}\right)^{-\frac{2}{3}}\times 9\times 64x^{9-1}$$

Whakarūnātia.

$$192x^{8}\times \left(64x^{9}\right)^{-\frac{2}{3}}$$