Problem 24 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 23 through 29, differentiate. In Problems 23 through 25, assume \(f\) is differentiable. Your answers may be in terms of \(f\) and \(f^{\prime} .\) $$ y=[f(x)]^{2}+2^{f(x)} $$

Step-by-Step Solution

Answer

The derivative of \(y= [f(x)]^{2} + 2^{f(x)}\) is \(y'= (2)f(x) f'(x) + 2^{f(x)} \ln(2) f'(x)\).

Step 1: Use the power rule on \([f(x)]^{2}\)

Recall that the power rule states that the derivative of \(x^n\) is \(n \cdot x^{n-1}\). Applying this rule, the derivative of \([f(x)]^{2}\) is \((2)f(x) f'(x)\), using the chain rule.

Step 2: Use the chain rule on \(2^{f(x)}\)

Next is to differentiate \(2^{f(x)}\). Recall that the derivative of \(a^u\) is \(a^u \ln(a) u'\) where \(a\) is a constant, \(u\) is a variable. Also recall that the natural logarithm of \(2\) is \(\ln(2)\), and is also a constant in this case. Therefore, the derivative of \(2^{f(x)}\) is \(2^{f(x)} \ln(2) f'(x)\).

Step 3: Combine the terms

Combining the calculated derivatives, which are the results from steps 1 and 2, we get the derivative of \(y\) i.e., \(y'= (2)f(x) f'(x) + 2^{f(x)} \ln(2) f'(x)\).

Key Concepts

Chain Rule

When you have a function within another function, the chain rule comes into play in differentiation. It helps find the derivative of such composite functions. In simpler terms, if you have something like \( g(h(x)) \), you apply the derivative of the outer function \( g \) while keeping the inner function \( h(x) \) unchanged at first. Multiply this by the derivative of the inner function \( h'(x) \).
For example, in the exercise, we differentiate \([f(x)]^2\) as part of the function \(y=[f(x)]^{2}+2^{f(x)}\). Here, your "inner function" is \(f(x)\), which is kept as is before multiplying by its derivative \( f'(x) \) after using the power rule.

Power Rule

The power rule is one of the fundamental rules of differentiation. It states that the derivative of \( x^n \) is \( n \cdot x^{n-1} \). This rule simplifies the process of finding derivatives of power functions where exponents are constants.
In our specific example, we're dealing with \([f(x)]^2\). Applying the power rule here means taking the exponent \( 2 \) out front, leaving us with \( 2 \cdot f(x) \). Don't forget, we also need to multiply it by \( f'(x) \) due to the chain rule since \( f(x) \) is itself a function that needs differentiation. So the derivative of \([f(x)]^2\) becomes \( (2)f(x)f'(x) \).
This approach is quick and efficient, making differentiation of polynomials straightforward.

Derivative of Exponential Functions

Exponential functions, like \( a^{f(x)} \), have their own rule for differentiation. When differentiating such functions, the derivative is \( a^{f(x)} \cdot \ln(a) \cdot f'(x) \). This accounts for both the exponential nature and the change in the function \( f(x) \).
In our case, we differentiate \( 2^{f(x)} \). The base here is \( 2 \), which remains untouched aside from adding a natural log term as part of the derivative: \( \ln(2) \). Don’t forget to multiply by the derivative \( f'(x) \) of the exponent function \( f(x) \). Thus, \( 2^{f(x)} \) transforms into \( 2^{f(x)} \ln(2) f'(x) \) when differentiated.
Understanding this helps enormously with all kinds of real-world exponential growth or decay problems.

Differentiable Function

A function is differentiable at a point if it has a defined derivative there, meaning you can tweak one variable slightly and predict how another changes. This smoothness (having no sharp corners or jumps) is a core requirement for differentiation, ensuring mathematical tools like the chain rule work effectively.
For the function \( f(x) \) in our exercise, assuming it's differentiable means we can find its derivative \( f'(x) \). This assumption holds the base for using all our derivative rules effectively. When a function meets the criteria of being continuous and smooth enough, it tells us that we can use derivative operations across its domain, making life much easier when applying complex calculus operations like those we've done here.