# Thread: derivative of sine function with exponent x

1. ## derivative of sine function with exponent x

Okay, so I've tried doing this problem a million times and I can't seem to get the right answer. It seems fairly straight forward.
Here is the question:
If f(x)=6(sin(x))x , find f(3)

So this is what I've tried:
lny=6xlnsinx
(1/y)(y')=f'g+fg'
(1/y)(y')=(6)(lnsinx)+[(6x)((cosx)/(sinx))]
y'=[(6lnsin(x))+(6xcot(x))]*(6(sin(x))^x)
So I tried pluggin in 3 to the answer, but it does not give me the correct answer.
Am I doing something wrong?

(I've also tried with the assumption that lny=xln6sinx, but that didn't work either.
Thanx!

2. There's no reason to take logs, it's just a product of functions, apply the product rule... or is it just that I'm misunderstanding the problem and it's $\displaystyle 6(\sin x)^x$ ?

3. Ya sorry, the question is actually f(x)=6(sin(x))^x like you said. Thought I typed it like that.. sorry...

4. Well in that case, it does make sense to apply logs, but make it easier, put $\displaystyle (\sin x)^x=e^{x\ln(\sin x)},$ now, contemplate its derivative.

5. uhhhh... sorry, I think you lost me. Where did you get that expression from?

6. Because of $\displaystyle x=e^{\ln x}$ for $\displaystyle x>0.$

7. okay, so I think I found the derivative.
I got:

[(lnsinx)+(xcotx)][6e^(xlnsinx)]
But it still isn't right.
Did I do this wrong?

8. Well, having $\displaystyle f(x)=6(\sin x)^x=6e^{x\ln(\sin x)},$ hence $\displaystyle f'(x)=6e^{x\ln(\sin x)}\cdot(x\ln(\sin x))',$ because of $\displaystyle (x\ln(\sin x))'=\ln(\sin x)+x\cot x$ the derivative equals $\displaystyle f'(x)=6(\sin x)^x\cdot(\ln(\sin x)+x\cot x)$ and we're done.

9. Thanks. That makes complete sense. I actually did get it right the last time, I was looking at the wrong answer in my textbook. Haha. Thanks for all your help!!