Problem 38 - Thomas Calculus

Question

Find the derivative of \(y\) with respect to the given independent variable. \(y=t \log _{3}\left(e^{(\sin t)(\ln 3)}\right)\)

Step-by-Step Solution

Answer

The derivative of \( y \) with respect to \( t \) is \( \sin t + t\cos t \).

Step 1: Simplify the Expression

First, focus on simplifying the expression inside the logarithm. Given the expression inside the logarithm as \( e^{(\sin t)(\ln 3)} \), we use the property of exponents which states that \( a^{bc} = (a^b)^c \). Hence, \( e^{(\sin t)(\ln 3)} = (e^{\ln 3})^{\sin t} = 3^{\sin t} \). Substitute back into the original equation to get \( y = t \log_{3}(3^{\sin t}) \).

Step 2: Apply Logarithmic Property

Use the property of logarithms which states \( \log_{b}(b^x) = x \). Thus, \( \log_{3}(3^{\sin t}) = \sin t \). Substitute this back into the equation for \( y \). Thus, \( y = t \sin t \).

Step 3: Differentiate the Simplified Expression

Now differentiate the expression \( y = t \sin t \) with respect to \( t \). Use the product rule of differentiation, which states that if \( u(t) = t \) and \( v(t) = \sin t \), then \( \frac{d}{dt}(uv) = u'v + uv' \). Derive as follows: \[ \frac{dy}{dt} = \frac{d}{dt}(t \sin t) = (1)(\sin t) + (t)(\cos t) = \sin t + t \cos t. \]

Step 4: Finalize the Derivative

Combine the terms if needed, but here we already have the derivative in its simplest form: \( \frac{dy}{dt} = \sin t + t\cos t \). This represents the rate of change of \( y \) with respect to \( t \).

Key Concepts

Logarithmic Functions

Logarithmic functions are fundamental in calculus and algebra, offering a way to reverse exponentiation. They have the form \(\log_{b}(x)\) where \(b\) is the base, and \(x\) is the argument. An essential property of logarithms is that they convert multiplication into addition, making complex calculations simpler.
Such properties include:

• \(\log_{b}(xy) = \log_{b}(x) + \log_{b}(y)\)
• \(\log_{b}(\frac{x}{y}) = \log_{b}(x) - \log_{b}(y)\)
• \(\log_{b}(x^c) = c\log_{b}(x)\)
• \(\log_{b}(b^x) = x\), where this specific property was used in the exercise to simplify the expression.

Using these properties, we can manipulate the logarithm for easier differentiation. In the exercise, converting \(e^{(\sin t)(\ln 3)}\) allowed simplification to \(3^{\sin t}\), which then applied the property \(\log_{3}(3^{\sin t}) = \sin t\), leading to a simplified function.

Product Rule

The Product Rule is a crucial tool in differentiation, particularly when dealing with functions expressed as products. When a function \(y = u(t) \cdot v(t)\) is given, where both \(u(t)\) and \(v(t)\) are functions of the same variable, you can differentiate it using the product rule.
The formula is:

• \(\frac{d}{dt}(uv) = u'v + uv'\)

Where \(u'\) is the derivative of \(u\) and \(v'\) is the derivative of \(v\).In the exercise, the product rule was applied to \( y = t \sin t \) by setting \( u(t) = t \) and \( v(t) = \sin t \), resulting in:

• \(u' = 1\) (derivative of \(t\))
• \(v' = \cos t\) (derivative of \(\sin t\))

Thus, \( \frac{d}{dt}(t \sin t) = (1\cdot \sin t) + (t \cdot \cos t) = \sin t + t\cos t\).Understanding how to apply the product rule is essential for differentiating similar expressions.

Exponential Functions

Exponential functions are vital, often appearing in growth and decay models in mathematics. These functions generally have the form \(f(t) = a \cdot b^t\), where \(a\) is a constant, \(b\) is the base of the exponential function, and \(t\) is the exponent. A specific and very useful property is that the derivative of an exponential function \(e^{x}\) with respect to \(x\) is itself, \(\frac{d}{dx}(e^x) = e^x\).
In problems involving exponential functions within a logarithm, like the example where \( e^{(\sin t)(\ln 3)} \) appears, they may simplify. The expression was rewritten using properties of exponents: converting it to a base familiar in logarithms, such as \(3\), aids in further simplification. This enabled the simplification steps in which an intermediate exponential \(3^{\sin t}\) was formed, making differentiation straightforward after converting it inside the logarithm.Knowledge of these fundamental principles helps solve more complex problems and recognize patterns within different mathematical contexts.