Problem 114 - Thomas Calculus in SI Units

Question

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=t^{\sqrt{t}}$$

Step-by-Step Solution

Answer

The derivative is \(\frac{dy}{dt} = t^{\sqrt{t} - \frac{1}{2}} (\frac{1}{2} \ln t + 1)\).

Step 1: Take the Natural Logarithm of Both Sides

Start with the equation \(y = t^{\sqrt{t}}\). Take the natural logarithm of both sides to transform the expression in a way that simplifies taking the derivative: \[ \ln y = \ln(t^{\sqrt{t}}) \]Use the logarithm power rule to bring the exponent down: \[ \ln y = \sqrt{t} \cdot \ln t \].

Step 2: Differentiate Implicitly with Respect to t

Differentiate both sides implicitly with respect to \(t\). Remember that \(y\) is a function of \(t\), so when differentiating \(\ln y\), you must use the chain rule:\[ \frac{d}{dt}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dt} \]For the right side, differentiate \(\sqrt{t} \cdot \ln t\) using the product rule: Let \(u = \sqrt{t}\) and \(v = \ln t\), then:\[ \frac{du}{dt} = \frac{1}{2\sqrt{t}} \quad \text{and} \quad \frac{dv}{dt} = \frac{1}{t} \]Applying the product rule:\[ \frac{d}{dt}(uv) = u'v + uv' = \frac{1}{2\sqrt{t}} \cdot \ln t + \sqrt{t} \cdot \frac{1}{t} \].

Step 3: Simplify the Right Side of the Equation

Compute each term of the differentiated expression:\[ \frac{1}{2\sqrt{t}} \cdot \ln t = \frac{\ln t}{2\sqrt{t}} \]\[ \sqrt{t} \cdot \frac{1}{t} = \frac{1}{\sqrt{t}} \]Combine these results:\[ \frac{dy}{dt} = y \left( \frac{\ln t}{2\sqrt{t}} + \frac{1}{\sqrt{t}} \right) \].

Step 4: Substitute y Back in Terms of t

Since we started with \(y = t^{\sqrt{t}}\), substitute back: \[ \frac{dy}{dt} = t^{\sqrt{t}} \left( \frac{\ln t}{2\sqrt{t}} + \frac{1}{\sqrt{t}} \right) \].

Step 5: Simplify the Expression

Factor out \(\frac{1}{\sqrt{t}}\) from the expression:\[ \frac{dy}{dt} = \frac{t^{\sqrt{t}}}{\sqrt{t}} (\frac{1}{2} \ln t + 1) \] Since \(\frac{t^{\sqrt{t}}}{\sqrt{t}} = t^{\sqrt{t} - \frac{1}{2}}\), the final simplified form becomes:\[ \frac{dy}{dt} = t^{\sqrt{t} - \frac{1}{2}} \left( \frac{1}{2} \ln t + 1 \right) \].

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used for finding the derivative of the composition of two or more functions. It's essential when dealing with complex derivatives, as it allows you to break down the differentiation process into more manageable steps.
When you have a composite function like \( f(g(t)) \), the chain rule states that the derivative \( \frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t) \). This seems a bit abstract, so let's look at an example from logarithmic differentiation.
In our problem, when differentiating \( \ln y \), remember that \( y \) is a function of \( t \). Therefore, we apply the chain rule: \( \frac{d}{dt}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dt} \).
This highlights how the chain rule helps us deal with implicit differentiation cases, where the function \( y \) depends on another variable \( t \). Right after using the chain rule, the derivative of \( y \) is expressed in terms of \( t \), allowing us to solve for \( \frac{dy}{dt} \) effectively.

Product Rule

The product rule is invaluable when differentiating products of functions. In situations where functions are multiplied together, like \( u(t) \cdot v(t) \), you can't just take the derivative of each and multiply. Instead, the product rule is the key.

• If \( y = u(t) \cdot v(t) \), then \( \frac{dy}{dt} = u'(t)v(t) + u(t)v'(t) \).

The solution we examined employs this rule while differentiating \( \sqrt{t} \cdot \ln t \).
To apply the product rule here, let \( u = \sqrt{t} \) and \( v = \ln t \). First, find the derivatives of \( u \) and \( v \):

• \( \frac{du}{dt} = \frac{1}{2\sqrt{t}} \)
• \( \frac{dv}{dt} = \frac{1}{t} \)

Then substitute these into the product rule formula:
\[ \frac{d}{dt}(uv) = \frac{1}{2\sqrt{t}} \cdot \ln t + \sqrt{t} \cdot \frac{1}{t} \].
This simplifies the differentiation process, helping us distinguish the individual contributions of each component in the function product.

Implicit Differentiation

Implicit differentiation is a technique used when a function is not given explicitly. In other words, when you don’t have a straightforward \( y = f(t) \) format. Sometimes, functions are intermingled with respect to variable \( t \), and we need implicit differentiation to find \( \frac{dy}{dt} \).
In this specific problem, after taking the natural logarithm of both sides, we differentiate explicitly with respect to \( t \), treating \( y \) as an implicit function of \( t \). Using implicit differentiation involves differentiating each term in the equation with respect to \( t \), simultaneously using the chain rule where necessary:
\[ \frac{d}{dt}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dt} \].
Implicit differentiation helps us to handle equations where separating variables may otherwise be challenging. It’s handy, especially when combined with logarithmic differentiation, as it simplifies the derivative process by reducing the power of variables.

Natural Logarithm

A natural logarithm, abbreviated as \( \ln \), is the logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. Natural logarithms have unique properties that simplify otherwise complex calculus tasks, such as exponent differentiation.

• Taking the natural logarithm of both sides helps by converting multiplication into addition and exponentiation into multiplication, making differentiation easier.

In our problem, by taking the logarithm, the original form \( y = t^{\sqrt{t}} \) transforms to \( \ln y = \sqrt{t} \cdot \ln t \), making differentiation straightforward through implicit techniques.
This transformation exploits properties like the logarithmic power rule: \( \ln(a^b) = b \cdot \ln a \), a critical step in making differentiation more feasible.
Understanding and utilizing such logarithmic properties is crucial when performing logarithmic differentiation. It's not just about simplifying equations, but also transforming them to a more "derivative-friendly" form.