What is the derivative of #y = x^cos(x)#?

2 Answers
Jan 15, 2017

#dy/dx = x^cosx(-sinxlnx + cosx/x)#

Explanation:

#y = x^cosx#

Take the natural logarithm of both sides.

#lny = ln(x^cosx)#

Use the logarithm law for powers, which states that #loga^n = nloga#

#lny = cosxlnx#

Use the product rule to differentiate the right hand side. #d/dx(cosx) = -sinx# and #d/dx(lnx)#.

#1/y(dy/dx) = -sinx(lnx) + cosx(1/x)#

#1/y(dy/dx) = -sinxlnx + cosx/x#

#dy/dx = (-sinxlnx + cosx/x)/(1/y)#

#dy/dx = x^cosx(-sinxlnx + cosx/x)#

Hopefully this helps!

Jan 15, 2017

#d/(dx) x^(cosx) = x^(cosx) (cosx/x -sinx lnx )#

Explanation:

You can write:

#x^(cosx) = (e^lnx)^(cosx) = e^(lnxcosx)#

so:

#d/(dx) x^(cosx) = d/(dx) (e^(lnxcosx)) #

using the chain rule:

#d/(dx) x^(cosx) = e^(lnxcosx) d/(dx) (lnxcosx)#

then the product rule:

#d/(dx) x^(cosx) = x^(cosx) (cosx/x -sinx lnx )#