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secondary 4 | A Maths
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Differentiation of trigonometry.
Anyone has idea for this question?
Thanks.
Date Posted: 8 months ago
Views: 121
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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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