Tayng's answer to Vincent Lim's Secondary 4 A Maths Singapore question.
done
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clear
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I think
Date Posted:
8 months ago
why "negative"?
d/dx [sin(2kx)] = k ⋅ [ -cos(2kx) ] ⋅ 2k
d/dx [sin(2kx)] = k ⋅ [ -cos(2kx) ] ⋅ 2k
Ohhhhhhhhhh oops
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