Whakaoti i te 5p^3q^2/25pq=

$$\frac { 5 p ^ { 3 } q ^ { 2 } } { 25 p q } =$$

Aromātai

$$\frac{qp^{2}}{5}$$

Ngā Upane e Whakamahi ana i te Āhuatanga o te Otinga o ngā Pū

Whakamahia ngā ture taupū hei whakarūnā i te kīanga.

$$\frac{5^{1}p^{3}q^{2}}{25^{1}p^{1}q^{1}}$$

Hei whakawehe i ngā pū o te pūtake kotahi, tangohia te taupū o te tauraro mai i te taupū o te taurunga.

$$\frac{5^{1}}{25^{1}}p^{3-1}q^{2-1}$$

Tango 1 mai i 3.

$$\frac{5^{1}}{25^{1}}p^{2}q^{2-1}$$

Tango 1 mai i 2.

$$\frac{5^{1}}{25^{1}}p^{2}q^{1}$$

Whakahekea te hautanga \frac{5}{25} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 5.

$$\frac{1}{5}p^{2}q$$

Kimi Pārōnaki e ai ki p

$$\frac{2pq}{5}$$

Ngā Upane Otinga

Hei whakawehe i ngā pū o te pūtake kotahi, tangohia te taupū o te tauraro mai i te taupū o te taurunga.

$$\frac{\mathrm{d}}{\mathrm{d}p}(\frac{5q^{2}}{25q}p^{3-1})$$

Mahia ngā tātaitanga.

$$\frac{\mathrm{d}}{\mathrm{d}p}(\frac{q}{5}p^{2})$$

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}.

$$2\times \frac{q}{5}p^{2-1}$$

Mahia ngā tātaitanga.

$$\frac{2q}{5}p^{1}$$

Mō tētahi kupu t, t^{1}=t.

$$\frac{2q}{5}p$$