I'm having a little trouble differentiating the function y=x^(x^0.5) or y=x^rootx
The answer in my math textbook is; [x^(rootx-1/2)][(1/2)lnx+1] and I can't seem to get this
Here are the steps I followed;
y=x^(rootx)
lny=(rootx)lnx
y'/y=(lnx/2rootx)+(rootx/x)
y'=y[(lnx/2rootx)+(rootx/x)]
y'=[x^(rootx)][(lnx/2rootx)+(rootx/x)]
At this point I can't seem to get any closer to the textbook given answer.
Any help would be greatly appreciated.
The answer in my math textbook is; [x^(rootx-1/2)][(1/2)lnx+1] and I can't seem to get this
Here are the steps I followed;
y=x^(rootx)
lny=(rootx)lnx
y'/y=(lnx/2rootx)+(rootx/x)
y'=y[(lnx/2rootx)+(rootx/x)]
y'=[x^(rootx)][(lnx/2rootx)+(rootx/x)]
At this point I can't seem to get any closer to the textbook given answer.
Any help would be greatly appreciated.
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Your work is perfect.
Here is another way:
Use e function,
y = e^[rootx lnx]
y' = [x^(rootx)][lnx/(2rootx) + rootx/x] = (1/2)[x^(rootx - (1/2))][lnx + 2]
Is this the answer in the textbook?
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Factor out 1/rootx = x^(-1/2)
Here is another way:
Use e function,
y = e^[rootx lnx]
y' = [x^(rootx)][lnx/(2rootx) + rootx/x] = (1/2)[x^(rootx - (1/2))][lnx + 2]
Is this the answer in the textbook?
---------
Factor out 1/rootx = x^(-1/2)