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\frac{\mathrm{d}}{\mathrm{d}x}(5x^{2}+1)
Me whakakore tahi te 5 i te taurunga me te tauraro.
2\times 5x^{2-1}
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}.
10x^{2-1}
Whakareatia 2 ki te 5.
10x^{1}
Tango 1 mai i 2.
10x
Mō tētahi kupu t, t^{1}=t.
5x^{2}+1
Me whakakore tahi te 5 i te taurunga me te tauraro.