Derivative of \(\displaystyle{x}^{{x}}\) without logarithmic differentiation With logarithmic

ropowiec2gkc

ropowiec2gkc

Answered question

2022-03-26

Derivative of xx without logarithmic differentiation
With logarithmic differentiation, it is quite simple to compute the derivative of xx:
y=xx
lny=xlnx
1ydydx=ln{x}+1
dydx=(ln(x)+1)xx.
Is there a method to compute the derivative of xx that does not rely on logarithmic differentiation?

Answer & Explanation

Carter Lin

Carter Lin

Beginner2022-03-27Added 13 answers

y={x}x={e}xln{x}
dydx=(ln(x)+1){e}xln{x}
dydx=(ln(x)+1){x}x
glikozyd3s68

glikozyd3s68

Beginner2022-03-28Added 16 answers

There's another way that looks like a gross blunder but actually is perfectly correct. I will illustrate it on the more general question, differentiate y=f(x)g(x)
If g were constant, we'd get f(x)g(x)f(x)g(x)1
If f were constant, we'd get g(x)f(x)g(x)logf(x)
Add these together to get the answer:
y=f(x)g(x)f(x)g(x)1+g(x)f(x)g(x)logf(x)
It's easy to see that in the original problem, where f(x)=x and g(x)=x, this reduces to xx+xxlogx , as obtained by other methods.

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