Whakaoti i te (5x^2+3x+6)+(7x^2-2x+3)

$$(5 { x }^{ 2 } +3x+6)+(7 { x }^{ 2 } -2x+3)$$

Aromātai

$$12x^{2}+x+9$$

Ngā Upane Otinga

Pahekotia te 5x^{2} me 7x^{2}, ka 12x^{2}.

$$12x^{2}+3x+6-2x+3$$

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

$$12x^{2}+x+6+3$$

Tāpirihia te 6 ki te 3, ka 9.

$$12x^{2}+x+9$$

Kimi Pārōnaki e ai ki x

$$24x+1$$

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

Pahekotia te 5x^{2} me 7x^{2}, ka 12x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(12x^{2}+3x+6-2x+3)$$

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

$$\frac{\mathrm{d}}{\mathrm{d}x}(12x^{2}+x+6+3)$$

Tāpirihia te 6 ki te 3, ka 9.

$$\frac{\mathrm{d}}{\mathrm{d}x}(12x^{2}+x+9)$$

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}.

$$2\times 12x^{2-1}+x^{1-1}$$

Whakareatia 2 ki te 12.

$$24x^{2-1}+x^{1-1}$$

Tango 1 mai i 2.

$$24x^{1}+x^{1-1}$$

Tango 1 mai i 1.

$$24x^{1}+x^{0}$$

Mō tētahi kupu t, t^{1}=t.

$$24x+x^{0}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$24x+1$$