What is the derivative of #x^sin(x)#?

1 Answer
Teddy
Jun 25, 2018

#d/dx(x^sinx)=x^sinx(cosxlogx+sinx/x)# where #logx# is the natural logarithm.

Explanation:

Let #y=x^sinx#. We will use logarithmic differentiation to calculate #dy/dx#.
#logy=log(x^sinx)#
#logy=sinxlogx#
Now, differentiating both sides with respect to #x#, we have
#d/dx(logy)=d/dx(sinxlogx)#

To compute the left side, remember the chain rule:
#y=f(g(x))iffdy/dx=f'(g(x))g'(x)#. Letting #f(x)=logx# and #g(x)=y#, the left side becomes #d/dx(logy)=1/ydy/dx#.
#:.1/ydy/dx=d/dx(sinxlogx)#

To differentiate #sinxlogx#, remember the product rule:
#y=f(x)g(x)iffdy/dx=f'(x)g(x)+g'(x)f(x)#
#:.1/ydy/dx=cosxlogx+sinx/x#

Multiplying through by #y#, we have
#dy/dx=y(cosxlogx+sinx/x)#
Remembering that we defined #y=x^sinx#, we get
#dy/dx=x^sinx(cosxlogx+sinx/x)#

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