Find the derivative of y with respect to x, where y = ln(1/(x(x+1))).
Understand the Problem
The question is asking us to find the derivative of the function y with respect to x, where y is defined as the natural logarithm of a fraction involving x. This involves using differentiation rules such as the chain rule and the quotient rule.
Answer
The derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = -\frac{2x + 1}{x(x + 1)} \).
Answer for screen readers
The derivative of ( y ) with respect to ( x ) is:
$$ \frac{dy}{dx} = -\frac{2x + 1}{x(x + 1)} $$
Steps to Solve
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Rewrite the function using properties of logarithms
We can use the property of logarithms that states $\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$.
Thus, we can rewrite the function ( y = \ln\left(\frac{1}{x(x + 1)}\right) ) as:
$$ y = \ln(1) - \ln(x(x + 1)) $$
Since $\ln(1) = 0$, we simplify to:
$$ y = -\ln(x(x + 1)) $$ -
Apply the chain rule
Next, we differentiate ( y ) with respect to ( x ). Using the chain rule, we find the derivative:
$$ \frac{dy}{dx} = -\frac{d}{dx}(\ln(x(x + 1))) $$ -
Differentiate using the product rule
We apply the product rule to differentiate ( x(x + 1) ):
$$ \frac{d}{dx}(x(x + 1)) = x\cdot\frac{d}{dx}(x + 1) + (x + 1)\cdot\frac{d}{dx}(x) = x(1) + (x + 1)(1) = 2x + 1 $$ -
Combine the previously obtained results
Now, substitute the derivative back into the chain rule:
$$ \frac{dy}{dx} = -\frac{1}{x(x + 1)}(2x + 1) $$ -
Final expression for the derivative
We simplify the expression to obtain:
$$ \frac{dy}{dx} = -\frac{2x + 1}{x(x + 1)} $$
The derivative of ( y ) with respect to ( x ) is:
$$ \frac{dy}{dx} = -\frac{2x + 1}{x(x + 1)} $$
More Information
This derivative shows how the function ( y ) changes with respect to ( x ). The negative sign indicates that the function decreases as ( x ) increases.
Tips
- Forgetting to apply the chain rule correctly when differentiating the logarithmic function.
- Misapplying the product rule, which can lead to incorrect derivatives.
- Neglecting to simplify the final derivative expression.
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