Whakaoti i te d/dxleft(3x^2-1/x-3right)

$$\frac{d}{d x } \left( \frac{ 3 { x }^{ 2 } -1 }{ x-3 } \right)$$

Aromātai

$$\frac{3x^{2}-18x+1}{\left(x-3\right)^{2}}$$

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

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.

$$\frac{\left(x^{1}-3\right)\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2}-1)-\left(3x^{2}-1\right)\frac{\mathrm{d}}{\mathrm{d}x}(x^{1}-3)}{\left(x^{1}-3\right)^{2}}$$

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}.

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

Mahia ngā tātaitanga.

$$\frac{\left(x^{1}-3\right)\times 6x^{1}-\left(3x^{2}-1\right)x^{0}}{\left(x^{1}-3\right)^{2}}$$

Whakarohaina mā te āhuatanga tohatoha.

$$\frac{x^{1}\times 6x^{1}-3\times 6x^{1}-\left(3x^{2}x^{0}-x^{0}\right)}{\left(x^{1}-3\right)^{2}}$$

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

$$\frac{6x^{1+1}-3\times 6x^{1}-\left(3x^{2}-x^{0}\right)}{\left(x^{1}-3\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{6x^{2}-18x^{1}-\left(3x^{2}-x^{0}\right)}{\left(x^{1}-3\right)^{2}}$$

Tangohia ngā taiapa kāore i te hiahiatia.

$$\frac{6x^{2}-18x^{1}-3x^{2}-\left(-x^{0}\right)}{\left(x^{1}-3\right)^{2}}$$

Pahekotia ngā kīanga tau ōrite.

$$\frac{\left(6-3\right)x^{2}-18x^{1}-\left(-x^{0}\right)}{\left(x^{1}-3\right)^{2}}$$

Tango 3 mai i 6.

$$\frac{3x^{2}-18x^{1}-\left(-x^{0}\right)}{\left(x^{1}-3\right)^{2}}$$

Mō tētahi kupu t, t^{1}=t.

$$\frac{3x^{2}-18x-\left(-x^{0}\right)}{\left(x-3\right)^{2}}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$\frac{3x^{2}-18x-\left(-1\right)}{\left(x-3\right)^{2}}$$

Kimi Pārōnaki e ai ki x

$$\frac{52}{\left(x-3\right)^{3}}$$