Whakaoti i te (7u^2-5u+3)+(-u^2+2u+4)

$$( 7 u ^ { 2 } - 5 u + 3 ) + ( - u ^ { 2 } + 2 u + 4 )$$

Aromātai

$$6u^{2}-3u+7$$

Ngā Upane Otinga

Pahekotia te -5u me 2u, ka -3u.

$$7u^{2}-3u+3-u^{2}+4$$

Tāpirihia te 3 ki te 4, ka 7.

$$7u^{2}-3u+7-u^{2}$$

Pahekotia te 7u^{2} me -u^{2}, ka 6u^{2}.

$$6u^{2}-3u+7$$

Kimi Pārōnaki e ai ki u

$$12u-3$$

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

Pahekotia te -5u me 2u, ka -3u.

$$\frac{\mathrm{d}}{\mathrm{d}u}(7u^{2}-3u+3-u^{2}+4)$$

Tāpirihia te 3 ki te 4, ka 7.

$$\frac{\mathrm{d}}{\mathrm{d}u}(7u^{2}-3u+7-u^{2})$$

Pahekotia te 7u^{2} me -u^{2}, ka 6u^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}u}(6u^{2}-3u+7)$$

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}.

$$2\times 6u^{2-1}-3u^{1-1}$$

Whakareatia 2 ki te 6.

$$12u^{2-1}-3u^{1-1}$$

Tango 1 mai i 2.

$$12u^{1}-3u^{1-1}$$

Tango 1 mai i 1.

$$12u^{1}-3u^{0}$$

Mō tētahi kupu t, t^{1}=t.

$$12u-3u^{0}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$12u-3$$