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 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 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 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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