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How do you differentiate #y=x^(x-1)#?

Answer 1

Use logarithmic differentiation to get #y'=(lnx+1-1/x)x^(x-1)#.

Take the natural logarithm of both sides, to drop the exponent: #lny=(x-1)lnx#

Now differentiate both sides with respect to #x#: #d/dx(lny=(x-1)lnx)#

Note that this will require knowledge of implicit differentiation, because the derivative of #lny# w.r.t.x is #1/y*y'#. The derivative of #(x-1)lnx# is found with the product rule: #d/dx((x-1)(lnx))=(x-1)'(lnx)+(x-1)(lnx)'# #=(1)(lnx)+(x-1)*(1/x)# #=lnx+(x-1)/x# #=lnx+1-1/x#

Since #d/dx(lny)=1/y*y'#, and #d/dx((x-1)(lnx))=lnx+1-1/x#, we have: #1/y*y'=lnx+1-1/x#

Multiply both sides by #y# to isolate #y'#: #y'=(lnx+1-1/x)y#

Since #y=x^(x-1)#: #y'=(lnx+1-1/x)x^(x-1)#

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Answer 2

Samantha Adkison

To differentiate ( y = x^{x-1} ), you can use the product rule along with the chain rule. The derivative is ( y' = x^{x-1} (1 + \ln(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.