g(t)= 4sec(t) + tan(t) when finding d/dx g(t), is the derivative of 4sec(t) 0(tan x) = 0? or am i doing that wrong? thanks in advance
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Originally Posted by ExCaLiBuR g(t)= 4sec(t) + tan(t) when finding d/dx g(t), is the derivative of 4sec(t) 0(tan x) = 0? or am i doing that wrong? thanks in advance I think you are getting confusing with something else. I don't see where you got the 0 from... $\displaystyle \frac{d}{dx}g(t) = 4\sec(t)\tan(t) + sec^2t$ You could factorize the sec(t) if you wanted to.
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