Problem 41 - Calculus - AP Edition

Question

Find \(d y / d x\) using any method. $$y=4^{3 \sin x-e^{x}}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = e^{(3 \sin x - e^x) \ln 4} \cdot (3 \ln 4 \cdot \cos x - e^x \ln 4) \).

Step 1: Express y using natural exponent

Rewrite the expression for \( y \) in terms of the natural exponent \( e \). We know \( a^b = e^{b \ln a} \). Therefore, express \( y = 4^{3 \sin x - e^x} \) as:\[ y = e^{(3 \sin x - e^x) \ln 4} \].

Step 2: Differentiate using the chain rule

Differentiate \( y = e^{(3 \sin x - e^x) \ln 4} \) with respect to \( x \). Use the chain rule: if \( y = e^u \), then \( \frac{dy}{dx} = e^u \cdot \frac{du}{dx} \). Thus, let \( u = (3 \sin x - e^x) \ln 4 \):\[ \frac{dy}{dx} = e^{(3 \sin x - e^x) \ln 4} \cdot \frac{d}{dx}((3 \sin x - e^x) \ln 4) \].

Step 3: Differentiate the exponent

Differentiate \( u = (3 \sin x - e^x) \ln 4 \) with respect to \( x \):- The derivative of \( 3 \sin x \ln 4 \) is \( 3 \ln 4 \cdot \cos x \).- The derivative of \( -e^x \ln 4 \) is \( -e^x \ln 4 \).Thus, \( \frac{du}{dx} = 3 \ln 4 \cdot \cos x - e^x \ln 4 \).

Step 4: Combine results

Combine the results from Step 2 and Step 3 to find \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = e^{(3 \sin x - e^x) \ln 4} \cdot (3 \ln 4 \cdot \cos x - e^x \ln 4) \]This is the expression for the derivative of \( y \) with respect to \( x \).

Key Concepts

Chain Rule

The chain rule is an essential technique for differentiation, especially when dealing with compositions of functions. It enables us to differentiate a function inside another function in a systematic way.

• The chain rule states: if you have a composite function \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is given by \( f'(g(x)) \cdot g'(x) \).
• This essentially means you first differentiate the outer function keeping the inner function unchanged, then multiply it by the derivative of the inner function.
• In our exercise, the outer function \( e^u \) contains the entire expression \((3 \sin x - e^x) \ln 4 \) as the inner function.
• By using the chain rule, we first differentiate \( e^u \) with respect to \( u \) and then the inner function with respect to \( x \).

This approach helps break down complex derivatives into manageable parts, making it easier to solve Differential Calculus problems.

Exponential Functions

Exponential functions are a fascinating class of functions that involve the constant \( e \), approximately 2.71828. They have a unique property that makes their usage widespread in calculus and other areas of mathematics.

• When differentiating \( e^x \), it gives itself back as the derivative, \( \frac{d}{dx}e^x = e^x \).
• This property allows them to grow at a rate proportional to their value, showcasing exponential growth.
• In our problem, we rewrite \( y = 4^{3 \sin x - e^x} \) as \( e^{(3 \sin x - e^x) \ln 4} \) using the identity \( a^b = e^{b \ln a} \).
• This transformation is crucial because it lets us apply differentiation rules easily, especially when paired with the known exponential function behavior.

Understanding exponential functions and their derivatives is beneficial for tackling various mathematical models involving growth and decay.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is another key concept often used in calculus problems involving exponential processes. It's the inverse function to the exponential function with base \( e \).

• The natural logarithm of a number \( a \) is the power to which \( e \) must be raised to get that number. Formally, if \( e^x = a \), then \( \ln a = x \).
• Within the differentiation process, logarithms help simplify expressions and are particularly useful in manipulating exponents.
• In the exercise, we used \( \ln 4 \) to transform the base 4 exponential expression into an exponential expression with base \( e \) — making it easier to differentiate.
• Differentiation rules involving \( \ln \) such as \( \frac{d}{dx} \ln(x) = \frac{1}{x} \) also come in handy during calculus problems, though they are not directly used in this specific exercise.

Having a clear understanding of natural logarithms can significantly enhance your problem-solving toolkit in calculus.