Whakaoti i te d/dx(2x^3-3x)^-4

$$\frac { d } { d x } ( 2 x ^ { 3 } - 3 x ) ^ { - 4 }$$

Aromātai

$$\frac{12\left(1-2x^{2}\right)}{\left(x\left(2x^{2}-3\right)\right)^{5}}$$

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).

$$-4\left(2x^{3}-3x^{1}\right)^{-4-1}\frac{\mathrm{d}}{\mathrm{d}x}(2x^{3}-3x^{1})$$

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}.

$$-4\left(2x^{3}-3x^{1}\right)^{-5}\left(3\times 2x^{3-1}-3x^{1-1}\right)$$

Whakarūnātia.

$$\left(2x^{3}-3x^{1}\right)^{-5}\left(-24x^{2}+12x^{0}\right)$$

Mō tētahi kupu t, t^{1}=t.

$$\left(2x^{3}-3x\right)^{-5}\left(-24x^{2}+12x^{0}\right)$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$\left(2x^{3}-3x\right)^{-5}\left(-24x^{2}+12\times 1\right)$$

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

$$\left(2x^{3}-3x\right)^{-5}\left(-24x^{2}+12\right)$$

Kimi Pārōnaki e ai ki x

$$\frac{12\left(52x^{4}-48x^{2}+15\right)}{\left(x\left(2x^{2}-3\right)\right)^{6}}$$