Question

Use logarithmic differentiation to find the derivative of the function. y= x^(4cosx)

Answer 1

Take ln on both sides.
Use lna^b = b*ln a
Differentiate left and right
Multiply with y

Answer 2

This is what I got from differentiating on both sides: (1/y)*y'=-4sinx*lnx+(1/x)*4cosx

Answer 3

good, now just go move your y to the numerator on the other side and change it back to what you started out with, which was y=x^(4cosx)

Answer 4

OK, so now multiply left and right with y to get y'

Answer 5

Yeah, what ZeHanz said.

Answer 6

And be sure to replace y with x^(4cosx), like @calmat01 said :)

Answer 7

lol..Thanks, @ZeHanz

Answer 8

I got y'=-4sin(x)*lnx+(1/x)*4cos(x)*x^(4cosx)

Answer 9

Make sure you put the brackets around the right terms. To be on the safe side, I would put the x^(4cosx) first and then put everything after it in one big bracket.

Answer 10

so it would be x^(4cosx)*[-4sin(x)*lnx+(1/x)*4cosx]?

Answer 11

There ya go.

Answer 12

Yes, though you could write 1/x*4cosx as \[\frac{ 4\cos x }{ x }\]

Answer 13

you could probably simplify that 1/x)..lol

Answer 14

lol Ya beat me to it, @ZeHanz

Answer 15

Yes, it's fun isn't it!

Answer 16

could the answer also be 4x^(4cos(x))*[cosx/(x)-ln(x)*sin(x)]?

Answer 17

Yes, it could. By factoring out a 4 from both terms.

Answer 18

awesome! thank you both for helping me through the problem :)

Answer 19

Np. Glad to be of help.

Answer 20

YW!