Whakaoti i te cos(3θ+2θ)

$$\cos ( 3 \theta +2 \theta )$$

Kimi Pārōnaki e ai ki θ

$$-5\sin(5\theta )$$

Ngā Upane e Whakamahi ana i te Ture Mekameka

Pahekotia te 3\theta me 2\theta , ka 5\theta .

$$\frac{\mathrm{d}}{\mathrm{d}\theta }(\cos(5\theta ))$$

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).

$$\left(-\sin(5\theta ^{1})\right)\frac{\mathrm{d}}{\mathrm{d}\theta }(5\theta ^{1})$$

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}.

$$\left(-\sin(5\theta ^{1})\right)\times 5\theta ^{1-1}$$

Whakarūnātia.

$$-5\sin(5\theta ^{1})$$

Mō tētahi kupu t, t^{1}=t.

$$-5\sin(5\theta )$$

Aromātai

$$\cos(5\theta )$$