given y=sec(x^2)ln(2x^5)
find dy/dx
Joe, you have asked 5 rather routine questions without showing us any effort on your part.
I will do this one, .. basic product rule
dy/dx = sec(x^2)(10x/(2x^5) + ln(2x^5)(sec(x^2)tanx^2)(2x)
= (5/x)sec(x^2) + 2x(ln(2x^5))sec(x^2)tan(x^2)
do the others using product rule, chain rule, and quotient rule, or combinations of those.
You will of course have to know the derivatives of the 6 trig ratios.