Problem 20 - Thomas Calculus

Question

Find the derivative of \(y\) with respect to the given independent variable. \(y=3^{\tan \theta} \ln 3\)

Step-by-Step Solution

Answer

\( \frac{dy}{d\theta} = 3^{\tan \theta} \ln 3 \cdot \sec^2 \theta \)

Step 1: Identify the Function Type

The function given is of the form \( y = a^{u(x)} \ln a \), which is a variation of the exponential form \( a^{u(x)} \) with additional \( \ln a \). Here, \( a = 3 \) and \( u(x) = \tan \theta \). This recognizes it as an exponential function with a variable exponent.

Step 2: Apply the Chain Rule

The derivative \( \frac{dy}{d\theta} \) is calculated using the chain rule for a function of the type \( y = a^{u(x)} \ln a \). The rule \( \frac{d}{dx}[a^{u(x)} \ln a] = a^{u(x)} \ln a \cdot u'(x) \) where \( u'(x) \) is the derivative of the exponent.So, we start with: \[ \frac{dy}{d\theta} = 3^{\tan \theta} \ln 3 \cdot \frac{d}{d\theta}(\tan \theta) \]

Step 3: Differentiate the Exponent

The exponent in the function is \( \tan \theta \). We need to find \( \frac{d}{d\theta}(\tan \theta) \). The derivative of \( \tan \theta \) with respect to \( \theta \) is \( \sec^2 \theta \).

Step 4: Substitute and Simplify

Substitute \( \frac{d}{d\theta}(\tan \theta) = \sec^2 \theta \) back into the derivative expression:\[ \frac{dy}{d\theta} = 3^{\tan \theta} \ln 3 \cdot \sec^2 \theta \]This gives you the final simplified derivative of the original function.

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where the variable appears in the exponent. This makes them unique, as they grow very rapidly compared to polynomial functions. In our exercise, the function is of the form \(y = 3^{\tan \theta} \ln 3\). Here, the base of the exponent, \(3\), is a constant, and the exponent function is \(\tan \theta\). Exponential functions are crucial in various fields like finance, physics, and biology, where they often model growth or decay.

• Standard Form: The standard exponential function is \(a^{x}\), where \(a\) is a positive constant, and \(x\) is any real number.
• Properties: These functions have a natural base, \(e\), leading to the natural exponential function \(e^{x}\), which is extensively used in calculus.

Understanding exponential functions is key to grasping more complex calculus problems where these functions are used alongside other operations, like derivatives.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. A composite function is a function composed of two or more functions, such as \(f(g(x))\). To differentiate such a function, you need to use the chain rule, which essentially states:

• If \(y = f(g(x))\), then the derivative \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\).

In our exercise, the chain rule helps to find \(\frac{d}{d\theta}[3^{\tan \theta} \ln 3]\). Here, \(3^{\tan \theta} \ln 3\) is a composite function where \(\tan \theta\) is nested within an exponential function. So, to differentiate, we multiply the derivative of the outer function by the derivative of the inner function:

• Outer Function: \(3^{u} \ln 3\) with \(u = \tan \theta\). The derivative is \(3^{\tan \theta} \ln 3 \cdot u'\).
• Inner Function: \(u = \tan \theta\), where \(u' = \sec^2 \theta\).

The result is smoothly linking these derivatives to find the overall derivative of the given function.

Trigonometric Derivatives

Trigonometric derivatives involve the differentiation of trigonometric functions like sine, cosine, and tangent. These derivatives are essential parts of many calculus problems, particularly involving physics and engineering.In this example, the derivative of \(\tan \theta\) is used. This derivative is central to solving our problem as it forms the inner function of the composite exponential function:

• \(\frac{d}{d\theta}\tan \theta = \sec^2 \theta\)

The derivative \(\sec^2 \theta\) arises frequently in calculus, often when dealing with integrals and derivatives of trigonometric functions. Knowing and applying these derivatives allows you to handle more complex equations and has practical utility in modeling periodic phenomena. By understanding how these derivatives interact, you can break down intricate problems into manageable parts, just like in the given exercise.