given y=sec(x^2)ln(2x^5)

find dy/dx

1 answer

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.
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