Whakaoti i te sin(2u)=

$$\sin ( 2 u ) =$$

Kimi Pārōnaki e ai ki u

$$2\cos(2u)$$

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).

$$\cos(2u^{1})\frac{\mathrm{d}}{\mathrm{d}u}(2u^{1})$$

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}.

$$\cos(2u^{1})\times 2u^{1-1}$$

Whakarūnātia.

$$2\cos(2u^{1})$$

Mō tētahi kupu t, t^{1}=t.

$$2\cos(2u)$$

Aromātai

$$\sin(2u)$$