Whakaoti i te (n+7)^1/2

$$( n + 7 ) ^ { 1 / 2 }$$

Kimi Pārōnaki e ai ki n

$$\frac{1}{2\sqrt{n+7}}$$

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).

$$\frac{1}{2}\left(n^{1}+7\right)^{\frac{1}{2}-1}\frac{\mathrm{d}}{\mathrm{d}n}(n^{1}+7)$$

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}.

$$\frac{1}{2}\left(n^{1}+7\right)^{-\frac{1}{2}}n^{1-1}$$

Whakarūnātia.

$$\frac{1}{2}n^{0}\left(n^{1}+7\right)^{-\frac{1}{2}}$$

Mō tētahi kupu t, t^{1}=t.

$$\frac{1}{2}n^{0}\left(n+7\right)^{-\frac{1}{2}}$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$\frac{1}{2}\times 1\left(n+7\right)^{-\frac{1}{2}}$$

Mō tētahi kupu t, t\times 1=t me 1t=t.

$$\frac{1}{2}\left(n+7\right)^{-\frac{1}{2}}$$

Aromātai

$$\sqrt{n+7}$$