Problem 24 - Thomas Calculus

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. \begin{equation}y=e^{\sin t}\left(\ln t^{2}+1\right)\end{equation}

Step-by-Step Solution

Answer

The derivative is \(\frac{dy}{dt} = e^{\sin t} \left(\cos t (\ln t^2 + 1) + \frac{2}{t}\right)\).

Step 1: Identify the function type

The given function is a product of two functions: \(e^{\sin t}\) and \(\ln t^2 + 1\). To find its derivative with respect to \(t\), we'll use the product rule for derivatives.

Step 2: Recall the Product Rule

The product rule states that the derivative of a product of two functions \(u(t)\) and \(v(t)\) is \((uv)' = u'v + uv'\). Here, we'll consider \(u = e^{\sin t}\) and \(v = \ln t^2 + 1\).

Step 3: Differentiate \(u = e^{\sin t}\)

Use the chain rule to find the derivative of \(u\). The derivative of \(e^{\sin t}\) is \(e^{\sin t} \cdot \cos t\) because \(\frac{d}{dt}(\sin t) = \cos t\).

Step 4: Differentiate \(v = \ln t^2 + 1\)

For \(v\), differentiate each part: \(\ln t^2\) using the chain rule gives \(\frac{1}{t^2}\cdot2t = \frac{2}{t}\), and the constant \(1\) has a derivative of \(0\). So, \(v' = \frac{2}{t}\).

Step 5: Apply the Product Rule

Substitute \(u'\), \(v\), \(u\), and \(v'\) into the product rule: \(\frac{dy}{dt} = (e^{\sin t} \cdot \cos t)(\ln t^2 + 1) + (e^{\sin t})(\frac{2}{t})\).

Step 6: Simplify the Expression

Combine like terms (if any) and write the expression neatly: \(\frac{dy}{dt} = e^{\sin t} \left(\cos t (\ln t^2 + 1) + \frac{2}{t}\right)\). This is the derivative of \(y\) with respect to \(t\).

Key Concepts

Product Rule

When faced with a function that is the product of two simpler functions, the product rule is indispensable. In calculus, the product rule helps us find the derivative of a product of two functions. The rule states that if you have two functions, say \( u(t) \) and \( v(t) \), the derivative of their product is given by:

• \( (uv)' = u'v + uv' \)

This means we differentiate one function while leaving the other one unchanged, then switch their roles. Finally, we add the results together. By applying this method, we manage to capture how both components of the product contribute to the overall rate of change.
It's akin to dealing with compound effects, like how two pedals on a bicycle work together to propel the bicycle forward. Understanding and applying the product rule is a fundamental skill in differentiation.

Chain Rule

The chain rule is a powerful tool for finding the derivative of composite functions. A composite function is one that is made by combining two or more functions, such as \( e^{\sin t} \) in the example, where \( \sin t \) is nested inside the exponential function.

• The chain rule formula is: \( \frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t) \)

When you're looking at \( e^{\sin t} \), you treat \(\sin t\) as the "inner function" and the exponential as the "outer function." Differentiate the outer function while keeping the inner function unchanged, and then multiply by the derivative of the inner function.
For \( e^{\sin t} \), the derivative is \( e^{\sin t} \cdot \cos t \), as we differentiate the sine component to get \( \cos t \). Practicing the chain rule helps you handle more complicated functions by breaking them down into simpler pieces.

Exponential Functions

Exponential functions are everywhere in mathematics, commonly expressed as \( e^x \), where \( e \) is approximately 2.718, a unique and important number. One of the key properties of exponential functions is how they differentiate. The derivative of \( e^x \) is particularly simple because it is \( e^x \) itself.

• In transformation, such as \( e^{\sin t} \), the chain rule plays a vital role.

The exponential \( e^{\sin t} \) requires applying the chain rule, acknowledging that the function inside the exponent can change. This behavior is crucial in calculus because exponential functions model a wide range of real-world phenomena, from population growth to radioactive decay.
Grasp the basic properties of exponential functions, and you are well-equipped to tackle more intricate derivatives involving these fascinating and far-reaching mathematical functions.

Logarithmic Differentiation

Logarithmic differentiation is an essential technique, especially when dealing with complicated products or when the function involves powers or logarithms themselves. It's particularly useful for simplifying the derivatives of expressions where one part is a power or a logarithmic expression, such as \( \ln t^2 \).

• Differentiation of \( \ln t^2 \) depends on the chain rule again, yielding \( \frac{2}{t} \).

The idea behind logarithmic differentiation is to take the natural log of a function and then differentiate, using the properties of logarithms to simplify the expression. This not only facilitates the differentiation process but can also make the algebra more manageable.
Understand when to apply logarithmic differentiation by identifying complicated formulas. It'll demystify sections of calculus that at first seem challenging, showing its powerful utility in simplifying intricate derivative calculations.