Whakaoti i te d/dx(x^6-8x^4)^10=

$$\frac { d } { d x } ( x ^ { 6 } - 8 x ^ { 4 } ) ^ { 10 } =$$

Aromātai

$$20\left(3x^{2}-16\right)\left(x^{2}-8\right)^{9}x^{39}$$

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).

$$10\left(x^{6}-8x^{4}\right)^{10-1}\frac{\mathrm{d}}{\mathrm{d}x}(x^{6}-8x^{4})$$

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}.

$$10\left(x^{6}-8x^{4}\right)^{9}\left(6x^{6-1}+4\left(-8\right)x^{4-1}\right)$$

Whakarūnātia.

$$\left(x^{6}-8x^{4}\right)^{9}\left(60x^{5}-320x^{3}\right)$$

Kimi Pārōnaki e ai ki x

$$3540\left(x^{2}-8\right)^{8}x^{42}-37920\left(x^{2}-8\right)^{8}x^{40}+99840\left(x^{2}-8\right)^{8}x^{38}$$