derivative of x to the power of x
Understand the Problem
The question is asking how to find the derivative of the function f(x) = x^x, which involves applying the rules of differentiation, particularly logarithmic differentiation.
Answer
The derivative of the function $f(x) = x^x$ is $$ f'(x) = x^x(\ln(x) + 1) $$
Answer for screen readers
The derivative of the function $f(x) = x^x$ is
$$ f'(x) = x^x(\ln(x) + 1) $$
Steps to Solve
- Rewriting the function using logarithmic differentiation
To differentiate $f(x) = x^x$, we can use logarithmic differentiation. First, take the natural logarithm of both sides:
$$ \ln(f(x)) = \ln(x^x) $$
This can be simplified using the properties of logarithms:
$$ \ln(f(x)) = x \ln(x) $$
- Differentiate both sides
Now we differentiate both sides with respect to $x$. Using the chain rule on the left side, we get:
$$ \frac{1}{f(x)} f'(x) = \frac{d}{dx}(x \ln(x)) $$
On the right side, we apply the product rule:
$$ \frac{d}{dx}(x \ln(x)) = \ln(x) + 1 $$
- Isolating $f'(x)$
Now, we substitute our right-hand derivative back into the equation:
$$ \frac{1}{f(x)} f'(x) = \ln(x) + 1 $$
Multiply both sides by $f(x)$ to isolate $f'(x)$:
$$ f'(x) = f(x)(\ln(x) + 1) $$
- Substituting back $f(x)$
Remember that $f(x) = x^x$. Now we substitute this back into our expression for $f'(x)$:
$$ f'(x) = x^x(\ln(x) + 1) $$
This gives us the derivative of the original function.
The derivative of the function $f(x) = x^x$ is
$$ f'(x) = x^x(\ln(x) + 1) $$
More Information
Logarithmic differentiation is a useful technique for finding derivatives of functions where the variable is both the base and the exponent, like in the case of $x^x$. This method simplifies the process and prevents complications involving higher powers.
Tips
- Not using logarithmic differentiation when necessary can lead to errors in computation.
- Forgetting to apply both the product rule and the chain rule can cause mistakes in the differentiation process.
