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

Question

Charles Anctil · Accepted Answer

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

#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!

Emily Ammerman · Answer

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

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 )#

Adam Adkins · Answer

undefined To find the derivative of y = x^cos(x), we can use the chain rule.

First, we differentiate the exponent and then multiply it by the derivative of the base function.

The derivative of cos(x) is -sin(x), and the derivative of x^u where u is a function of x is given by u' * x^(u-1).

So, applying the chain rule, the derivative of y = x^cos(x) is:

dy/dx = cos(x) * x^(cos(x) - 1) - x^cos(x) * ln(x) * sin(x)