Whakaoti i te (sqrt{(2-x)^2+25})^2

$$( \sqrt { ( 2 - x ) ^ { 2 } + 25 } ) ^ { 2 }$$

Aromātai

$$\left(2-x\right)^{2}+25$$

Ngā Upane Otinga

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(2-x\right)^{2}.

$$\left(\sqrt{4-4x+x^{2}+25}\right)^{2}$$

Tāpirihia te 4 ki te 25, ka 29.

$$\left(\sqrt{29-4x+x^{2}}\right)^{2}$$

Tātaihia te \sqrt{29-4x+x^{2}} mā te pū o 2, kia riro ko 29-4x+x^{2}.

$$29-4x+x^{2}$$

Kimi Pārōnaki e ai ki x

$$2\left(x-2\right)$$

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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(2-x\right)^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(\sqrt{4-4x+x^{2}+25}\right)^{2})$$

Tāpirihia te 4 ki te 25, ka 29.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\left(\sqrt{29-4x+x^{2}}\right)^{2})$$

Tātaihia te \sqrt{29-4x+x^{2}} mā te pū o 2, kia riro ko 29-4x+x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(29-4x+x^{2})$$

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}.

$$-4x^{1-1}+2x^{2-1}$$

Tango 1 mai i 1.

$$-4x^{0}+2x^{2-1}$$

Tango 1 mai i 2.

$$-4x^{0}+2x^{1}$$

Mō tētahi kupu t, t^{1}=t.

$$-4x^{0}+2x$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$-4+2x$$