How do you differentiate #y = x^(2 cos x)#?

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Answer 1 · 2016-11-06T16:36:17

The derivative for this is found by the chain rule.

Let, #y=f(x)=x^(2cosx)#.

#:.dy/dx=d/(dx)(x^(2cosx))*d/(dx)(2cosx)#.

#:.dy/dx=(2cosx*x^(2cosx-1))*(-2sinx)#.

#:.dy/dx=-4sinxcosx*x^(2cosx-1)#. (answer).

Answer 2 · 2016-11-06T16:44:51

Use some version of logarithmic differentiation.

My current preferred form for logarithmic differfentiation is to rewrite as #e# to a power.

#y = x^(2cosx) = e^(2cosxlnx)#

Now use #d/dx(e^u) = e^u (du)/dx# to get

#dy/dx = e^(2cosxlnx) * d/dx(2cosxlnx)#

# = x^(2cosx) *[(2cosx)/x - 2sinxlnx]#

Answer 3 · 2016-11-06T17:27:23

Please see the explanation for the steps leading to:

#dy/dx = 2(cos(x)/x - sin(x)ln(x))x^(2cos(x))#

Let #ln(y) = ln(x^(2cos(x)))#

#ln(y) = 2cos(x)ln(x)#

#1/y(dy/dx) = -2sin(x)ln(x) + 2cos(x)/x#

#dy/dx = 2(cos(x)/x - sin(x)ln(x))y#

Substitute #x^(2cos(x))# for y:

#dy/dx = 2(cos(x)/x - sin(x)ln(x))x^(2cos(x))#