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

Question

Please help. Thank You,

Answer 1

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 simple

Answer 2

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.

Answer 3

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)]