How do you figure out the derivative of y=x^2tan2x?

Please help. Thank You,How do you figure out the derivative of y=x^2tan2x?
If y = u^v, where u and v are functions of x, then





y' = u^v ln(u) v' + v u^(v -1) u' = u^(v -1) [u ln(u) v' + v u']





Here, u = x and v = 2tan(2x). Therefore,





u' = 1 and v' = 2 sec^2(2x) (2) = 4 sec^2(2x). So,





y' = (x^(2tan(2x) -1)) [4x ln(x) sec^2(2x) + 2tan(2x)]





OBS. I assume you mean y = x^(2tan(2x)), that is, 2tan(2x) is the exponent of x. Otherwise, it's kinda simpleHow do you figure out the derivative of y=x^2tan2x?
f(x)=x^[2tan(2x)]





1. step use the natural logarithm





ln f(x)=ln {x^[2tan(2x)]}





ln x^y=y*ln x





therefore:





ln f(x)= [2tan(2x)]*ln x





2. step use the product rule and the chain rule for the differentiation





[1/f(x)]*f `(x)= u`(x)*v(x) + u(x)*v`(x)





u(x)=ln x


u`(x)=1/x


v(x)=2*tan(2x)


v`(x)=2*2*1/cos^2(2x)





[1/f(x)]*f `(x)= (1/x)*2*tan(2x) + ln x *4/cos^2(2x)





f `(x) = f(x) * (1/x)*2*tan(2x) + ln x *4/cos^2(2x)





f `(x) = x^[2tan(2x)] * [(1/x)*2*tan(2x) + ln x *4/cos^2(2x)]
use the property of log or ln like this: ln y= 2 tan2x ln x. since the x is raise by 2tan2x. then get the derivative of both sides like this: 1/y y'=2(sec2xtan2x)(2)(1/x). then combine like this: y'=y(4sec2xtan2x/x). i'm not sure of the answer. but its the one i know. hope it helps you.

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