Whakaoti i te 8x^3+3x-1x^2-2x+38x^3-1

$$8 x ^ { 3 } + 3 x - 1 x ^ { 2 } - 2 x + 38 x ^ { 3 } - 1$$

Aromātai

$$46x^{3}-x^{2}+x-1$$

Ngā Upane Otinga

Pahekotia te 3x me -2x, ka x.

$$8x^{3}+x-x^{2}+38x^{3}-1$$

Pahekotia te 8x^{3} me 38x^{3}, ka 46x^{3}.

$$46x^{3}+x-x^{2}-1$$

Kimi Pārōnaki e ai ki x

$$138x^{2}-2x+1$$

Ngā Upane e Whakamahi ana i te Ture Pārōnaki mō te Tapeke

Pahekotia te 3x me -2x, ka x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(8x^{3}+x-x^{2}+38x^{3}-1)$$

Pahekotia te 8x^{3} me 38x^{3}, ka 46x^{3}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(46x^{3}+x-x^{2}-1)$$

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}.

$$3\times 46x^{3-1}+x^{1-1}+2\left(-1\right)x^{2-1}$$

Whakareatia 3 ki te 46.

$$138x^{3-1}+x^{1-1}+2\left(-1\right)x^{2-1}$$

Tango 1 mai i 3.

$$138x^{2}+x^{1-1}+2\left(-1\right)x^{2-1}$$

Tango 1 mai i 1.

$$138x^{2}+x^{0}+2\left(-1\right)x^{2-1}$$

Whakareatia 1 ki te 1.

$$138x^{2}+x^{0}-2x^{2-1}$$

Tango 1 mai i 2.

$$138x^{2}+x^{0}-2x^{1}$$

Mō tētahi kupu t, t^{1}=t.

$$138x^{2}+x^{0}-2x$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$138x^{2}+1-2x$$