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

Answer 1

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

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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Answer 2

Luke Alvarez

To find the derivative of ( x^{\sin(x)} ), you can use the chain rule. The derivative is ( x^{\sin(x)} (\sin(x) \ln(x) + \cos(x)/x) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.