Whakaoti i te (n+9)+(4n^3+2n^2+4n+2)

$$( n + 9 ) + ( 4 n ^ { 3 } + 2 n ^ { 2 } + 4 n + 2 )$$

Aromātai

$$4n^{3}+2n^{2}+5n+11$$

Ngā Upane Otinga

Pahekotia te n me 4n, ka 5n.

$$5n+9+4n^{3}+2n^{2}+2$$

Tāpirihia te 9 ki te 2, ka 11.

$$5n+11+4n^{3}+2n^{2}$$

Kimi Pārōnaki e ai ki n

$$12n^{2}+4n+5$$

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

Pahekotia te n me 4n, ka 5n.

$$\frac{\mathrm{d}}{\mathrm{d}n}(5n+9+4n^{3}+2n^{2}+2)$$

Tāpirihia te 9 ki te 2, ka 11.

$$\frac{\mathrm{d}}{\mathrm{d}n}(5n+11+4n^{3}+2n^{2})$$

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}.

$$5n^{1-1}+3\times 4n^{3-1}+2\times 2n^{2-1}$$

Tango 1 mai i 1.

$$5n^{0}+3\times 4n^{3-1}+2\times 2n^{2-1}$$

Whakareatia 3 ki te 4.

$$5n^{0}+12n^{3-1}+2\times 2n^{2-1}$$

Tango 1 mai i 3.

$$5n^{0}+12n^{2}+2\times 2n^{2-1}$$

Whakareatia 3 ki te 4.

$$5n^{0}+12n^{2}+4n^{2-1}$$

Tango 1 mai i 2.

$$5n^{0}+12n^{2}+4n^{1}$$

Mō tētahi kupu t, t^{1}=t.

$$5n^{0}+12n^{2}+4n$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$5\times 1+12n^{2}+4n$$

Mō tētahi kupu t, t\times 1=t me 1t=t.

$$5+12n^{2}+4n$$