Whakaoti i te d/dxleft(1/3x^2right)

$$\frac{d}{d x } \left( \frac{ 1 }{ 3 { x }^{ 2 } } \right)$$

Aromātai

$$-\frac{2}{3x^{3}}$$

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).

$$-\left(3x^{2}\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2})$$

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}.

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

Whakarūnātia.

$$-6x^{1}\times \left(3x^{2}\right)^{-2}$$

Mō tētahi kupu t, t^{1}=t.

$$-6x\times \left(3x^{2}\right)^{-2}$$

Kimi Pārōnaki e ai ki x

$$\frac{2}{x^{4}}$$