Question

Evaluate the following derivatives. $$\frac{d}{d x}\left((x+1)^{2 x}\right)$$

Short answer

Question: Find the derivative of the function $$f(x)=(x+1)^{2x}$$ with respect to x. Answer: The derivative of the function $$f(x)=(x+1)^{2x}$$ with respect to x is $$\frac{d}{d x}\left((x+1)^{2 x}\right) = (x+1)^{2x}\ln(x+1)$$.

Step-by-step solution

Identify the Inner and Outer Functions

We have the function $$f(x)=(x+1)^{2x}$$. Here, we can see there are two functions: 1. Inner function: $$u(x)=x+1$$ 2. Outer function: $$g(u)=u^{2x}$$

Apply the Chain Rule

The Chain Rule states that: $$\frac{d[f(g(x))]}{dx}=\frac{df(g(x))}{dg(x)}\cdot\frac{dg(x)}{dx}$$ To apply this rule, first, we need to find the derivatives of the inner $$u(x)$$ and outer $$g(u)$$ functions.

Find the Derivative of the Inner Function

To find the derivative of $$u(x)=x+1$$, we take the derivative with respect to x: $$\frac{du(x)}{dx}=\frac{d(x+1)}{dx}=1$$

Find the Derivative of the Outer Function

To find the derivative of $$g(u)=u^{2x}$$, we use the Exponential Function Rule, which states that: $$\frac{d}{dx}a^{x}=a^{x}\ln(a)$$ In this case, $$a=u(x)=(x+1)$$. So, we have: $$\frac{dg(u)}{dx}=(x+1)^{2x}\ln(x+1)$$ But we also need to multiply it with $$\frac{du(x)}{dx}=1$$, as per the chain rule. So, the final expression is: $$\frac{dg(u)}{dx}=(x+1)^{2x}\ln(x+1)$$

Combine the Results Using the Chain Rule

Now, we can combine the results using the chain rule: $$\frac{d}{d x}\left((x+1)^{2 x}\right) = \frac{dg(u)}{dx} \cdot \frac{du(x)}{dx} = (x+1)^{2x}\ln(x+1) \cdot 1 = (x+1)^{2x}\ln(x+1)$$ So, the derivative of the function $$f(x)=(x+1)^{2x}$$ with respect to x is: $$\frac{d}{d x}\left((x+1)^{2 x}\right) = (x+1)^{2x}\ln(x+1)$$

Key concepts

Derivative of Exponential Functions

Understanding the derivative of exponential functions is crucial for many areas in calculus. Exponential functions are in the form of \( f(x) = a^x \), where \( a \) is a constant, and they represent continuous growth or decay. When taking the derivative of an exponential function, we rely on a special rule called the Exponential Function Rule. This rule is stated as \( \frac{d}{dx} a^{x} = a^{x} \ln(a) \), where \( \ln(a) \) is the natural logarithm of \( a \). In applying this rule, it's important to recognize the function's base—\( a \)—as this directly influences the derivative.

For example, in the case of \( (x+1)^{2x} \), we treat \( x+1 \) as our base \( a \). Therefore, the derivative will involve \( (x+1)^{2x} \ln(x+1) \), accurately reflecting the exponential growth rate while accounting for the changing base as the value of \( x \) changes.

Calculus

Calculus is a vast field of mathematics that deals with change and motion through derivatives and integrals. One of the fundamental concepts in calculus is the derivative, which measures how a function \( f(x) \) changes as \( x \) changes. This concept is essential for understanding rates of change in physics, economics, and other sciences.

• Derivatives help us find slopes of tangent lines to curves.
• They are used to optimize functions to find maximum or minimum values.
• Calculus also helps in computing areas under curves and the accumulated quantities through integrals.

Through the use of limits, calculus allows us to study these changes even when they become infinitely small, providing precise and powerful tools for analysis.

Chain Rule

The Chain Rule is a critical technique in calculus for finding the derivative of composite functions. A composite function can be thought of as a function within another function, resembling a chain of functions linked together. The Chain Rule is formally expressed as \( \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) \), where \( f \) and \( g \) are functions, and \( f'(g(x)) \) and \( g'(x) \) are their derivatives.

Applying the Chain Rule requires identifying the inner and outer functions—usually denoted as \( u(x) \) and \( g(u) \)—and then calculating their individual derivatives before multiplying them. This rule is essential when dealing with functions like \( (x+1)^{2x} \) because it involves an inner function \( x+1 \) and an outer function \( u^{2x} \). By breaking down complex functions into simpler parts, the Chain Rule simplifies the process of differentiation and makes it manageable.

Exponential Function Rule

The Exponential Function Rule simplifies the differentiation of functions in the form of \( a^{g(x)} \), where \( a \) is a constant base and \( g(x) \) is an exponent that is also a function of \( x \). This rule is an extension of the Exponential Rule mentioned earlier but specifically addresses situations where the exponent itself is variable. In such cases, the derivative is found by multiplying the original function by the natural logarithm of the base and the derivative of the exponent.

The formula is given by \( \frac{d}{dx}a^{g(x)} = a^{g(x)} \ln(a)g'(x) \). It is commonly used in conjunction with the Chain Rule when the exponent is more complex than a simple \( x \), like in our example where the exponent is \( 2x \). Understanding this rule is fundamental for students as it appears frequently in scenarios involving exponential growth and decay, such as in biology, economics, and physics.