Whakaoti i te k^80*k^-28:k^93

$$k ^ { 80 } \cdot k ^ { - 28 } : k ^ { 93 }$$

Aromātai

$$\frac{1}{k^{41}}$$

Ngā Upane Otinga

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 80 me te -28 kia riro ai te 52.

$$\frac{k^{52}}{k^{93}}$$

Tuhia anō te k^{93} hei k^{52}k^{41}. Me whakakore tahi te k^{52} i te taurunga me te tauraro.

$$\frac{1}{k^{41}}$$

Kimi Pārōnaki e ai ki k

$$-\frac{41}{k^{42}}$$

Ngā Upane e Whakamahi ana i te Ture Mekameka

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 80 me te -28 kia riro ai te 52.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{k^{52}}{k^{93}})$$

Tuhia anō te k^{93} hei k^{52}k^{41}. Me whakakore tahi te k^{52} i te taurunga me te tauraro.

$$\frac{\mathrm{d}}{\mathrm{d}k}(\frac{1}{k^{41}})$$

Mēnā ko F te hanganga o ngā pānga e rua e taea ana te pārōnaki f\left(u\right) me u=g\left(x\right), arā, mēnā ko F\left(x\right)=f\left(g\left(x\right)\right), ko te pārōnaki o F te pārōnaki o f e ai ki u whakareatia te pārōnaki o g e ai ki x, arā, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

$$-\left(k^{41}\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}k}(k^{41})$$

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}.

$$-\left(k^{41}\right)^{-2}\times 41k^{41-1}$$

Whakarūnātia.

$$-41k^{40}\left(k^{41}\right)^{-2}$$