Whakaoti i te {left(1-xright)}^frac{3{2}}

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

Kimi Pārōnaki e ai ki x

$$-\frac{3\sqrt{1-x}}{2}$$

Ngā Upane e Whakamahi ana i te Ture Mekameka

Mēnā ko F te hanganga o ngā pānga e rua e taea ana te pārōnaki f\left(u\right) me u=g\left(x\right), arā, mēnā ko F\left(x\right)=f\left(g\left(x\right)\right), ko te pārōnaki o F te pārōnaki o f e ai ki u whakareatia te pārōnaki o g e ai ki x, arā, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

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

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}.

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

Whakarūnātia.

$$-\frac{3}{2}x^{0}\sqrt{-x^{1}+1}$$

Mō tētahi kupu t, t^{1}=t.

$$-\frac{3}{2}x^{0}\sqrt{-x+1}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$-\frac{3}{2}\sqrt{-x+1}$$

Aromātai

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