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Kimi Pārōnaki e ai ki x
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\frac{\mathrm{d}}{\mathrm{d}x}(\sin(12x)+x+4-x-4)
Hei kimi i te tauaro o x+4, kimihia te tauaro o ia taurangi.
\frac{\mathrm{d}}{\mathrm{d}x}(\sin(12x)+4-4)
Pahekotia te x me -x, ka 0.
\frac{\mathrm{d}}{\mathrm{d}x}(\sin(12x))
Tangohia te 4 i te 4, ka 0.
\cos(12x^{1})\frac{\mathrm{d}}{\mathrm{d}x}(12x^{1})
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).
\cos(12x^{1})\times 12x^{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}.
12\cos(12x^{1})
Whakarūnātia.
12\cos(12x)
Mō tētahi kupu t, t^{1}=t.