Whakaoti i te cos2x(30+4)

$$\cos 2 x ( 30 + 4 )$$

Kimi Pārōnaki e ai ki x

$$-68\sin(68x)$$

Ngā Upane e Whakamahi ana i te Ture Mekameka

Tāpirihia te 30 ki te 4, ka 34.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\cos(2x\times 34))$$

Whakareatia te 2 ki te 34, ka 68.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\cos(68x))$$

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).

$$\left(-\sin(68x^{1})\right)\frac{\mathrm{d}}{\mathrm{d}x}(68x^{1})$$

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}.

$$\left(-\sin(68x^{1})\right)\times 68x^{1-1}$$

Whakarūnātia.

$$-68\sin(68x^{1})$$

Mō tētahi kupu t, t^{1}=t.

$$-68\sin(68x)$$

Aromātai

$$\cos(68x)$$