Whakaoti i te d/dz(z+3/2z-4)

$$\frac { d } { d z } ( \frac{z+3}{2z-4} )$$

Aromātai

$$-\frac{5}{2\left(z-2\right)^{2}}$$

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.

$$\frac{\left(2z^{1}-4\right)\frac{\mathrm{d}}{\mathrm{d}z}(z^{1}+3)-\left(z^{1}+3\right)\frac{\mathrm{d}}{\mathrm{d}z}(2z^{1}-4)}{\left(2z^{1}-4\right)^{2}}$$

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}.

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

Mahia ngā tātaitanga.

$$\frac{\left(2z^{1}-4\right)z^{0}-\left(z^{1}+3\right)\times 2z^{0}}{\left(2z^{1}-4\right)^{2}}$$

Whakarohaina mā te āhuatanga tohatoha.

$$\frac{2z^{1}z^{0}-4z^{0}-\left(z^{1}\times 2z^{0}+3\times 2z^{0}\right)}{\left(2z^{1}-4\right)^{2}}$$

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

$$\frac{2z^{1}-4z^{0}-\left(2z^{1}+3\times 2z^{0}\right)}{\left(2z^{1}-4\right)^{2}}$$

Mahia ngā tātaitanga.

$$\frac{2z^{1}-4z^{0}-\left(2z^{1}+6z^{0}\right)}{\left(2z^{1}-4\right)^{2}}$$

Tangohia ngā taiapa kāore i te hiahiatia.

$$\frac{2z^{1}-4z^{0}-2z^{1}-6z^{0}}{\left(2z^{1}-4\right)^{2}}$$

Pahekotia ngā kīanga tau ōrite.

$$\frac{\left(2-2\right)z^{1}+\left(-4-6\right)z^{0}}{\left(2z^{1}-4\right)^{2}}$$

Tangohia te 2 i 2 me te 6 i te -4.

$$\frac{-10z^{0}}{\left(2z^{1}-4\right)^{2}}$$

Mō tētahi kupu t, t^{1}=t.

$$\frac{-10z^{0}}{\left(2z-4\right)^{2}}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$\frac{-10}{\left(2z-4\right)^{2}}$$

Kimi Pārōnaki e ai ki z

$$\frac{5}{\left(z-2\right)^{3}}$$