Whakaoti i te tan(4x)

$$\tan ( 4x )$$

Kimi Pārōnaki e ai ki x

$$\frac{4\left(16\left(\sin(x)\right)^{2}\left(\cos(x)\right)^{6}+16\left(\cos(x)\right)^{2}\left(\sin(x)\right)^{6}+2\left(\sin(2x)\right)^{4}-4\left(\sin(2x)\right)^{2}+1\right)}{4\left(\sin(2x)\right)^{4}-4\left(\sin(2x)\right)^{2}+1}$$

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).

$$\left(\sec(4x^{1})\right)^{2}\frac{\mathrm{d}}{\mathrm{d}x}(4x^{1})$$

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}.

$$\left(\sec(4x^{1})\right)^{2}\times 4x^{1-1}$$

Whakarūnātia.

$$4\left(\sec(4x^{1})\right)^{2}$$

Mō tētahi kupu t, t^{1}=t.

$$4\left(\sec(4x)\right)^{2}$$

Aromātai

$$\tan(4x)$$