Take the derivative of f(x)=ln(sec√(x))
The answer is f '(x)=tan√(x)/ 2√(x) but I don't know how to get to it :(
The answer is f '(x)=tan√(x)/ 2√(x) but I don't know how to get to it :(
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f(x) = ln(sec√(x)),
f '(x) = [1/sec√(x)]*sec√(x)*tan√(x)*[1/2√(x)],
f '(x) = tan√(x) / 2√(x) >====================< ANSWER
f '(x) = [1/sec√(x)]*sec√(x)*tan√(x)*[1/2√(x)],
f '(x) = tan√(x) / 2√(x) >====================< ANSWER
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f(x)=ln(sec√(x)) = - ln(cos√(x)) , use chain rule ---> f '(x) = tan√(x)/ (2√(x))