Whakaoti i te (sqrt{x-1}-1)(sqrt{x-1}+1)

$$( \sqrt { x - 1 } - 1 ) ( \sqrt { x - 1 } + 1 )$$

Aromātai

$$x-2$$

Ngā Upane Otinga

Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\sqrt{x-1}\right)^{2}-1^{2}$$

Tātaihia te \sqrt{x-1} mā te pū o 2, kia riro ko x-1.

$$x-1-1^{2}$$

Tātaihia te 1 mā te pū o 2, kia riro ko 1.

$$x-1-1$$

Tangohia te 1 i te -1, ka -2.

$$x-2$$

Kimi Pārōnaki e ai ki x

$$1$$

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

Whakaarohia te \left(\sqrt{x-1}-1\right)\left(\sqrt{x-1}+1\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Tātaihia te \sqrt{x-1} mā te pū o 2, kia riro ko x-1.

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

Tātaihia te 1 mā te pū o 2, kia riro ko 1.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x-1-1)$$

Tangohia te 1 i te -1, ka -2.

$$\frac{\mathrm{d}}{\mathrm{d}x}(x-2)$$

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}.

$$x^{1-1}$$

Tango 1 mai i 1.

$$x^{0}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$1$$