Whakaoti i te h(x)=-x/3x^2-9

$$h ( x ) = \frac { - x } { 3 x ^ { 2 } - 9 }$$

Kimi Pārōnaki e ai ki x

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

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.

$$\frac{\left(3x^{2}-9\right)\frac{\mathrm{d}}{\mathrm{d}x}(-x^{1})-\left(-x^{1}\frac{\mathrm{d}}{\mathrm{d}x}(3x^{2}-9)\right)}{\left(3x^{2}-9\right)^{2}}$$

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}.

$$\frac{\left(3x^{2}-9\right)\left(-1\right)x^{1-1}-\left(-x^{1}\times 2\times 3x^{2-1}\right)}{\left(3x^{2}-9\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{\left(3x^{2}-9\right)\left(-1\right)x^{0}-\left(-x^{1}\times 6x^{1}\right)}{\left(3x^{2}-9\right)^{2}}$$

Whakarohaina mā te āhuatanga tohatoha.

$$\frac{3x^{2}\left(-1\right)x^{0}-9\left(-1\right)x^{0}-\left(-x^{1}\times 6x^{1}\right)}{\left(3x^{2}-9\right)^{2}}$$

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

$$\frac{3\left(-1\right)x^{2}-9\left(-1\right)x^{0}-\left(-6x^{1+1}\right)}{\left(3x^{2}-9\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{-3x^{2}+9x^{0}-\left(-6x^{2}\right)}{\left(3x^{2}-9\right)^{2}}$$

Pahekotia ngā kīanga tau ōrite.

$$\frac{\left(-3-\left(-6\right)\right)x^{2}+9x^{0}}{\left(3x^{2}-9\right)^{2}}$$

Tango -6 mai i -3.

$$\frac{3x^{2}+9x^{0}}{\left(3x^{2}-9\right)^{2}}$$

Tauwehea te 3.

$$\frac{3\left(x^{2}+3x^{0}\right)}{\left(3x^{2}-9\right)^{2}}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$\frac{3\left(x^{2}+3\times 1\right)}{\left(3x^{2}-9\right)^{2}}$$

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

$$\frac{3\left(x^{2}+3\right)}{\left(3x^{2}-9\right)^{2}}$$

Aromātai

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