find y' y=x^[cos(x)]
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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.
Originally Posted by Asuhuman18 find y' y=x^[cos(x)] logarithmic differentiation ... take the log of both sides $\displaystyle \ln{y} = \cos{x} \cdot \ln{x}$ now take the derivative and solve for y'
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
Originally Posted by Asuhuman18 so it would be 10e^[ln(x)*cos(x)] 10e^[ln(x)*cos(x)]* -sin/x $\displaystyle u = \cos (x) \cdot \ln (x)$ 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).
Originally Posted by mr fantastic $\displaystyle u = \cos (x) \cdot \ln (x)$ 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)?
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.
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