Find the Derivative of f(x) = x^log(x) using Logarithmic Differentiation

In summary, the formula for differentiating f(x)=x^log(x) is f'(x) = (ln(x) + 1)x^log(x). Yes, f(x)=x^log(x) can be simplified to f(x) = x^(1+log(x)). To differentiate f(x)=x^log(x) using the product rule, the formula f'(x) = u'v + uv' would be used, where u = x and v = x^log(x). The domain of f(x)=x^log(x) is all positive real numbers and the range is also all positive real numbers.
  • #1
Fairy111
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0

Homework Statement


Find the derivative of the following function f.


Homework Equations



f(x)=x^log(x)

The Attempt at a Solution



I really don't know how to do this, as it's x to the power of a function of x.
 
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  • #2


Use the fact that for any function u(x),

[tex]u(x) = e^{\ln(u(x))}[/tex]
 
  • #3


Or (almost the same thing) use "logarithmic differentiation".

If y= xlog x then log(y)= log(xlog(x))= log(x)log(x).

Differentiate both sides with respect to x, using the chain rule on the left side, and then solve for dy/dx.
 

What is the formula for differentiating f(x)=x^log(x)?

The formula for differentiating f(x)=x^log(x) is f'(x) = (ln(x) + 1)x^log(x).

Can f(x)=x^log(x) be simplified?

Yes, f(x)=x^log(x) can be simplified to f(x) = x^(1+log(x)).

How do you differentiate f(x)=x^log(x) using the product rule?

To differentiate f(x)=x^log(x) using the product rule, you would use the formula f'(x) = u'v + uv', where u = x and v = x^log(x). This would give you the final result of f'(x) = (ln(x) + 1)x^log(x).

What is the domain of f(x)=x^log(x)?

The domain of f(x)=x^log(x) is all positive real numbers.

What is the range of f(x)=x^log(x)?

The range of f(x)=x^log(x) is all positive real numbers.

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