Whakaoti i te d/dx(0.1(400-40x+x^2))

$$\frac { d } { d x } ( 0.1 ( 400 - 40 x + x ^ { 2 } ) )$$

Aromātai

$$\frac{x}{5}-4$$

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

Whakamahia te āhuatanga tohatoha hei whakarea te 0.1 ki te 400-40x+x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(40-4x+0.1x^{2})$$

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}.

$$-4x^{1-1}+2\times 0.1x^{2-1}$$

Tango 1 mai i 1.

$$-4x^{0}+2\times 0.1x^{2-1}$$

Whakareatia 2 ki te 0.1.

$$-4x^{0}+0.2x^{2-1}$$

Tango 1 mai i 2.

$$-4x^{0}+0.2x^{1}$$

Mō tētahi kupu t, t^{1}=t.

$$-4x^{0}+0.2x$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$-4+0.2x$$

Kimi Pārōnaki e ai ki x

$$0.2$$