Whakaoti i te 1/1+x

$$\frac { 1 } { 1 + x }$$

Kimi Pārōnaki e ai ki x

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

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).

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

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}.

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

Whakarūnātia.

$$-x^{0}\left(x^{1}+1\right)^{-2}$$

Mō tētahi kupu t, t^{1}=t.

$$-x^{0}\left(x+1\right)^{-2}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

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

Aromātai

$$\frac{1}{x+1}$$