Whakaoti i te (x/2+2/x^2)^prime

$$( \frac { x } { 2 } + \frac { 2 } { x ^ { 2 } } ) ^ { \prime }$$

Aromātai

$$\frac{1}{2}-\frac{4}{x^{3}}$$

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

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Ko te taurea pātahi iti rawa o 2 me x^{2} ko 2x^{2}. Whakareatia \frac{x}{2} ki te \frac{x^{2}}{x^{2}}. Whakareatia \frac{2}{x^{2}} ki te \frac{2}{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{xx^{2}}{2x^{2}}+\frac{2\times 2}{2x^{2}})$$

Tā te mea he rite te tauraro o \frac{xx^{2}}{2x^{2}} me \frac{2\times 2}{2x^{2}}, me tāpiri rāua mā te tāpiri i ō raua taurunga.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{xx^{2}+2\times 2}{2x^{2}})$$

Mahia ngā whakarea i roto o xx^{2}+2\times 2.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x^{3}+4}{2x^{2}})$$

Mō ngā pānga e rua e taea ana te pārōnaki, ko te pārōnaki o te otinga o ngā pānga e rua ko te tauraro whakareatia ki te pārōnaki o te taurunga tango i te taurunga whakareatia ki te pārōnaki o te tauraro, ā, ka whakawehea te katoa ki te tauraro kua pūruatia.

$$\frac{2x^{2}\frac{\mathrm{d}}{\mathrm{d}x}(x^{3}+4)-\left(x^{3}+4\right)\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2})}{\left(2x^{2}\right)^{2}}$$

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}.

$$\frac{2x^{2}\times 3x^{3-1}-\left(x^{3}+4\right)\times 2\times 2x^{2-1}}{\left(2x^{2}\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{2x^{2}\times 3x^{2}-\left(x^{3}+4\right)\times 4x^{1}}{\left(2x^{2}\right)^{2}}$$

Whakarohaina mā te āhuatanga tohatoha.

$$\frac{2x^{2}\times 3x^{2}-\left(x^{3}\times 4x^{1}+4\times 4x^{1}\right)}{\left(2x^{2}\right)^{2}}$$

Hei whakarea pū o te pūtake ōrite, tāpiri ana taupū.

$$\frac{2\times 3x^{2+2}-\left(4x^{3+1}+4\times 4x^{1}\right)}{\left(2x^{2}\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{6x^{4}-\left(4x^{4}+16x^{1}\right)}{\left(2x^{2}\right)^{2}}$$

Tangohia ngā taiapa kāore i te hiahiatia.

$$\frac{6x^{4}-4x^{4}-16x^{1}}{\left(2x^{2}\right)^{2}}$$

Pahekotia ngā kīanga tau ōrite.

$$\frac{\left(6-4\right)x^{4}-16x^{1}}{\left(2x^{2}\right)^{2}}$$

Tango 4 mai i 6.

$$\frac{2x^{4}-16x^{1}}{\left(2x^{2}\right)^{2}}$$

Tauwehea te 2x.

$$\frac{2x\left(x^{3}-8x^{0}\right)}{\left(2x^{2}\right)^{2}}$$

Hei hiki i te hua o ngā tau e rua, neke atu rānei ki tētahi pū, hīkina ia tau ki te pū ka tuhi ko tāna hua.

$$\frac{2x\left(x^{3}-8x^{0}\right)}{2^{2}\left(x^{2}\right)^{2}}$$

Hīkina te 2 ki te pū 2.

$$\frac{2x\left(x^{3}-8x^{0}\right)}{4\left(x^{2}\right)^{2}}$$

Hei hiki pū ki tētahi pū anō, me whakarea ngā taupū.

$$\frac{2x\left(x^{3}-8x^{0}\right)}{4x^{2\times 2}}$$

Whakareatia 2 ki te 2.

$$\frac{2x\left(x^{3}-8x^{0}\right)}{4x^{4}}$$

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

$$\frac{2\left(x^{3}-8x^{0}\right)}{4x^{4-1}}$$

Tango 1 mai i 4.

$$\frac{2\left(x^{3}-8x^{0}\right)}{4x^{3}}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$\frac{2\left(x^{3}-8\times 1\right)}{4x^{3}}$$

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

$$\frac{2\left(x^{3}-8\right)}{4x^{3}}$$

Kimi Pārōnaki e ai ki x

$$\frac{12}{x^{4}}$$