Math Problem Statement

In Exercises 45–58, use logarithmic differentiation to find the derivative of y with respect to the given independent variable. y = sqrt(t / (t + 1))

Solution

We are tasked with finding the derivative of the function y=tt+1y = \sqrt{\frac{t}{t+1}} using logarithmic differentiation. Let's break this down step by step:

Step 1: Rewrite the equation

We begin by expressing yy in a more convenient form for differentiation: y=(tt+1)1/2y = \left( \frac{t}{t+1} \right)^{1/2} This allows us to work more easily with logarithms.

Step 2: Take the natural logarithm of both sides

Using logarithmic differentiation, we apply the natural logarithm to both sides of the equation: lny=ln((tt+1)1/2)\ln y = \ln \left( \left( \frac{t}{t+1} \right)^{1/2} \right) By the power rule for logarithms, we bring the exponent 12\frac{1}{2} in front: lny=12ln(tt+1)\ln y = \frac{1}{2} \ln \left( \frac{t}{t+1} \right)

Step 3: Apply the logarithmic property for division

Using the property ln(ab)=lnalnb\ln \left( \frac{a}{b} \right) = \ln a - \ln b, we get: lny=12(lntln(t+1))\ln y = \frac{1}{2} \left( \ln t - \ln (t+1) \right)

Step 4: Differentiate both sides with respect to tt

We now differentiate both sides of the equation with respect to tt. On the left-hand side, we apply the chain rule: 1ydydt=12(1t1t+1)\frac{1}{y} \frac{dy}{dt} = \frac{1}{2} \left( \frac{1}{t} - \frac{1}{t+1} \right) The right-hand side is simplified as: 12(1t1t+1)=12(t+1)tt(t+1)=121t(t+1)\frac{1}{2} \left( \frac{1}{t} - \frac{1}{t+1} \right) = \frac{1}{2} \cdot \frac{(t+1) - t}{t(t+1)} = \frac{1}{2} \cdot \frac{1}{t(t+1)}

Step 5: Solve for dydt\frac{dy}{dt}

Now, multiply both sides by yy to solve for dydt\frac{dy}{dt}: dydt=y121t(t+1)\frac{dy}{dt} = y \cdot \frac{1}{2} \cdot \frac{1}{t(t+1)} Substitute the original expression for yy back into the equation: dydt=tt+1121t(t+1)\frac{dy}{dt} = \sqrt{\frac{t}{t+1}} \cdot \frac{1}{2} \cdot \frac{1}{t(t+1)}

Final Answer:

Thus, the derivative of yy with respect to tt is: dydt=12t(t+1)3/2t\frac{dy}{dt} = \frac{1}{2} \cdot \frac{\sqrt{t}}{(t+1)^{3/2} t}

Let me know if you'd like further clarification or more details on any steps!


Here are some related questions for further practice:

  1. Can you apply logarithmic differentiation to y=tty = t^t?
  2. How does logarithmic differentiation help with products of functions?
  3. What is the derivative of y=ln(t2+1t+2)y = \ln \left( \frac{t^2 + 1}{t+2} \right)?
  4. How would you differentiate a more complex function like y=(t3+1)1/3t21y = \frac{(t^3 + 1)^{1/3}}{t^2 - 1}?
  5. How does the chain rule apply to logarithmic differentiation?

Tip: Always look for opportunities to simplify your expressions before differentiating. This can reduce errors and make the differentiation process easier.

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Math Problem Analysis

Mathematical Concepts

Logarithmic Differentiation
Derivative
Algebra

Formulas

y = sqrt(t / (t + 1)) = (t / (t + 1))^(1/2)
ln(y) = ln((t / (t + 1))^(1/2))
d(ln(y))/dt = (1/2) * (1/t - 1/(t+1))
dy/dt = y * (1/2) * 1/(t(t + 1))

Theorems

Power Rule
Chain Rule
Logarithmic Properties

Suitable Grade Level

College Level Calculus