h(x) = sin2xcos2x

h'(x) = -4cos2xsin2x <--- doubt this is correct

--------------------

f(x) = cotx/sinx (unsure on how to do this....)

-thx

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- Feb 21st 2010, 11:42 AMmaybnxtseasntrig derivative help...
h(x) = sin2xcos2x

h'(x) = -4cos2xsin2x <--- doubt this is correct

--------------------

f(x) = cotx/sinx (unsure on how to do this....)

-thx - Feb 21st 2010, 11:58 AMione
- Feb 21st 2010, 12:02 PMskeeter