Kimi Pārōnaki e ai ki x
2\cos(2x+5)
Aromātai
\sin(2x+5)
Graph
Tohaina
Kua tāruatia ki te papatopenga
\cos(2x^{1}+5)\frac{\mathrm{d}}{\mathrm{d}x}(2x^{1}+5)
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(2x^{1}+5)\times 2x^{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}.
2\cos(2x^{1}+5)
Whakarūnātia.
2\cos(2x+5)
Mō tētahi kupu t, t^{1}=t.
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