# Thread: yet another painful derivative

1. ## yet another painful derivative

find y'

y=x^[cos(x)]

2. Originally Posted by Asuhuman18
find y'

y=x^[cos(x)]
Use the same approach as explained here: http://www.mathhelpforum.com/math-he...rivatives.html

If you need more help, please show your work and say where you get stuck.

3. Originally Posted by Asuhuman18
find y'

y=x^[cos(x)]
logarithmic differentiation ...

take the log of both sides

$\ln{y} = \cos{x} \cdot \ln{x}$

now take the derivative and solve for y'

4. Originally Posted by mr fantastic
Use the same approach as explained here: http://www.mathhelpforum.com/math-he...rivatives.html

If you need more help, please show your work and say where you get stuck.
so it would be 10e^[ln(x)*cos(x)]
10e^[ln(x)*cos(x)]* -sin/x

5. Originally Posted by Asuhuman18
so it would be 10e^[ln(x)*cos(x)]
10e^[ln(x)*cos(x)]* -sin/x
$u = \cos (x) \cdot \ln (x)$ and you have to use the product rule to differentiate this (the derivative you got for it is wrong).

And note that in your final answer you should re-write 10e^[ln(x)*cos(x)] in its original form of x^cos(x).

6. Originally Posted by mr fantastic
$u = \cos (x) \cdot \ln (x)$ and you have to use the product rule to differentiate this (the derivative you got for it is wrong).

And note that in your final answer you should re-write 10e^[ln(x)*cos(x)] in its original form of x^cos(x).
so is the final answer is x^[cos(x)]*-sin(x)*ln(x))+cos(x)*(1/x)?

7. Originally Posted by Asuhuman18
so is the final answer is x^[cos(x)]*(-sin(x)*ln(x))+cos(x)*(1/x))?
You will need the red brackets.