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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 3.R.67

Higher-order derivatives Find and simplify y''.
y = 2^x x

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First, identify the function y = 2^x * x. This is a product of two functions, so we will use the product rule to find the first derivative y'.
Apply the product rule: If y = u * v, then y' = u' * v + u * v'. Here, let u = 2^x and v = x. Find the derivatives u' and v'.
Calculate u': The derivative of 2^x with respect to x is 2^x * ln(2).
Calculate v': The derivative of x with respect to x is 1.
Substitute u', v, u, and v' into the product rule formula to find y'. Then, differentiate y' again to find y'', applying the product rule and chain rule as necessary. Simplify the expression for y''.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Higher-Order Derivatives

Higher-order derivatives refer to the derivatives of a function taken multiple times. The first derivative, denoted as f'(x), gives the rate of change of the function, while the second derivative, f''(x), provides information about the curvature or concavity of the function. Understanding how to compute these derivatives is essential for analyzing the behavior of functions in calculus.
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Product Rule

The product rule is a fundamental differentiation technique used when finding the derivative of a product of two functions. If u(x) and v(x) are two differentiable functions, the product rule states that the derivative of their product is given by u'v + uv'. This rule is crucial for differentiating functions like y = 2^x * x, where both components depend on x.
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Exponential Functions

Exponential functions are functions of the form f(x) = a^x, where a is a constant and x is the variable. These functions have unique properties, such as their derivatives being proportional to the function itself. In the context of the given function y = 2^x * x, recognizing the behavior of the exponential component is vital for accurately computing its derivatives.
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