If x^y = e^(-y), prove that dy/dx = log(x)/(1 + log(x))^2.

Understand the Problem

The question is asking to prove a relationship in calculus involving derivatives and logarithms. Specifically, it requires demonstrating that the derivative of a function defined by the equation x^y = e^(-y) matches a stated formula.

Answer

$$ \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} $$

Steps to Solve

Taking the logarithm of both sides
We start with the equation ( x^y = e^{-y} ). Taking the natural logarithm of both sides, we can write:
$$ \log(x^y) = \log(e^{-y}) $$
Using properties of logarithms, this becomes:
$$ y \log x = -y $$
Rearranging the equation
Next, we can rearrange the equation:
$$ y \log x + y = 0 $$
Factoring out ( y ) gives us:
$$ y(\log x + 1) = 0 $$
Thus:
$$ y = \frac{x}{1 + \log x} $$
Differentiating both sides with respect to x
Now, we differentiate ( y ) with respect to ( x ):
Using the quotient rule, where ( u = x ) and ( v = 1 + \log x ), we have:
$$ \frac{dy}{dx} = \frac{(v)(\frac{du}{dx}) - (u)(\frac{dv}{dx})}{v^2} $$
This translates to:
$$ \frac{dy}{dx} = \frac{(1 + \log x)(1) - (x)(\frac{1}{x})}{(1 + \log x)^2} $$
Simplifying the derivative
Now simplify:
$$ \frac{dy}{dx} = \frac{1 + \log x - 1}{(1 + \log x)^2} = \frac{\log x}{(1 + \log x)^2} $$

The derivative is given by
$$ \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} $$

More Information

The relationship derived shows that the derivative of ( y ) with respect to ( x ) involves the natural logarithm, showcasing the interplay between exponential and logarithmic functions. This result is useful in various applications, including growth and decay problems.

Tips

Forgetting to apply the product or quotient rule correctly: Ensure to recognize when to apply these rules while differentiating.
Neglecting to simplify: Always simplify your derivatives as it might lead to clearer results where conclusions can easily be drawn.

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