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Find the derivative: y = 2sinxtanx Is it: 2cosxtanx + sec^2(x)sinx Anyone mind checking for me?
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did you miss the 2 in the 2nd term?
It's supposed to be 2sinx because of the original?
yes
So, it would be 2cos(x)tan(x) + sec^2(x)2sin(x), yes?
2cos(x)tan(x) + 2sec^2(x)sin(x) would be convenient
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right which is same as 2cos(x)tan(x) + 2sec^2(x)sin(x) although this can be simplied further, but i don't think thats required.
Or would the 2 be throughout like this: 2cos(x)tan(x) + 2sec^2(x)sin(x)? (sorry my internet dropped)
Oh okay, I got it, thank you so much.
welcome ^_^
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