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

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Answer 1 · 2018-03-08T13:35:53

The answer is #=(1/xcosx-lnxsinx)x^(cosx)#

The function is

#y=x^(cosx)#

Taking logarithms on both sides

#lny=ln(x^(cosx))=cosxlnx#

Differentiating both sides

#(lny)'=(cosxlnx)'=(cosx)'lnx+cosx(lnx)'#

#1/ydy/dx=-sinx*lnx+cosx*1/x#

#dy/dx=(1/xcosx-lnxsinx)y#

#=(1/xcosx-lnxsinx)x^(cosx)#

Answer 2 · 2018-03-08T13:42:20

#x^cosx(cosx/x-sinxlnx)#

We know that #a^b=e^(blna)#. Here, we say that #x^cosx=e^(cosxlnx)#.

According to the chain rule, #(df)/dx=(df)/(du)*(du)/dx#, where #u# is a function within #f#.

Here, #f=e^u# where #u=cosxlnx#, so we have:

#d/(du)e^u*d/dx(cosxlnx)#

#e^ud/dx(cosxlnx)#

According to the product rule, #(f*g)'=f'g+fg'#. Here, #f=cosx# and #g=lnx#, so we have:

#d/dxcosx*lnx+d/dxlnx*cosx#

#-sinxlnx+1/xcosx#

#cosx/x-sinxlnx#

So we have:

#e^u(cosx/x-sinxlnx)#, but as #u=cosxlnx#, we have:

#e^(cosxlnx)(cosx/x-sinxlnx)#, but remember that #e^(cosxlnx)=x^cosx#, so we have:

#x^cosx(cosx/x-sinxlnx)#