# Differentiate y = (x^x) cos2x?

I solved it and arrived at (-2x^x)(sin2x) + (x^x)cos2x(ln(x) +1). Am i even the right track? how do I know what answer to choose among the given?

### 3 Answers

- stanschimLv 72 months ago
Use logarithmic differentiation:

ln(y)= ln[(x^x)(cos(2x))]

ln(y) = ln(x^x) + ln|cos(2x)|

ln(y) = xlnx + ln|cos(2x)|

Differentiate:

(1/y)(dy/dx) = x(1/x) + ln(x) + (1/cos(2x))(-2sin(2x))

(1/y)(dy/dx) = 1 + ln(x) - 2tan(2x)

dy/dx = y[1 + ln(x) - 2tan(2x)]

dy/dx = (x^x)(cos(2x))[1 + ln(x) - 2tan(2x)]

dy/dx = (x^x)(cos(2x))(1 + ln(x)) - 2(x^x)sin(2x))

Looks like the same answer you came up with. I question the choices given.

- AlanLv 72 months ago
None of the given answers are correct.

Your answer appears to be correct.

symbolab comes up with the same answer as you.

https://www.symbolab.com/solver/implicit-derivativ...

To check if your equation is equivalent to

other equations, just check a few values

like

0.1

1

4

However, none of the equation match your equation.

Something is wrong with your textbook or course.

- billrussell42Lv 72 months ago
rules:

(x^x)' = (x^x)(1 + ln x) (cos ax)' = –a sin ax product rule (fg)' = f'g + fg' y = (x^x) cos2xf = x^xf'= (x^x)(1 + ln x) g = cos2xg' = –2sin2xy' = (x^x)(1 + ln x)(cos2x) + (x^x)(–2sin2x)y' = (x^x)(1 + ln x)(cos2x) – 2(x^x)(sin2x)y' = (x^x)(cos2x + (ln x)(cos2x)) – 2(x^x)(sin2x)y' = (x^x)(cos2x + (ln x)(cos2x) – 2sin2x)I agree with your answer, don't see how any of the given answers are correct.