Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ y^{\prime} \text { if } y=e^{2 \ln x} $$

Step-by-Step Solution

Answer

The derivative is \( y' = 2x \).

Step 1: Simplify the Expression

Start by simplifying the expression for \( y \). Given that \( y = e^{2 \, \ln x} \), use the property of logarithms, \( a \, \ln b = \ln(b^a) \), to rewrite it. This gives \( y = e^{\ln(x^2)} \), which simplifies to \( y = x^2 \) because \( e^{\ln(a)} = a \).

Step 2: Differentiate the Simplified Expression

Now that you have simplified \( y = x^2 \), differentiate this expression with respect to \( x \). The derivative of \( x^2 \) with respect to \( x \) is found using the power rule: \( \frac{d}{dx}(x^n) = n \cdot x^{n-1} \). Applying this, you get \( y' = 2x \).

Key Concepts

Understanding Derivatives

In calculus, a derivative represents the rate at which a function changes. It is a fundamental concept that helps us understand how a small change in input affects the output of a function. When you see something like \( y' \) or \( \frac{dy}{dx} \), it denotes the derivative of \( y \) with respect to \( x \).

• The derivative represents the slope of the tangent line at any point on a curve.
• In practical terms, it's like understanding the speed (derivative) at a particular moment from the distance covered (function).

To find the derivative, different rules and techniques can be applied, depending on the function's form. Simplifying the expression before finding its derivative can often make the process easier. Once you grasp the essence of derivatives, solving problems involving them becomes more intuitive.

The Role of Logarithms

Logarithms are essential tools in mathematics, especially useful for simplifying complex expressions involving exponents. An important property of logarithms is that they can transform multiplication into addition, aiding in the simplification of expressions. For instance, if we have \( a \ln b = \ln(b^a) \), this property transforms the product of \( a \) and the logarithm of \( b \) into the logarithm of \( b \) raised to the power of \( a \).

• This property was used in our problem to transform \( e^{2 \ln x} \) into \( e^{\ln(x^2)} \).
• Subsequently, it simplified to \( x^2 \), as the expression \( e^{\ln a} \) simplifies to \( a \).

By understanding these properties, mathematicians can navigate through complex exponential expressions and simplify them to a more manageable form to differentiate.

Applying the Power Rule

The power rule is one of the most straightforward techniques to find the derivative of a function involving powers of \( x \). It states that for any function \( f(x) = x^n \), the derivative is \( f'(x) = n \cdot x^{n-1} \).

• In simple terms, bring the exponent down as a coefficient and subtract one from the original exponent.
• This rule was applied in the final step of our problem where the function was simplified to \( y = x^2 \).

So, using the power rule, we found that \( y' = 2x \), efficiently calculating the derivative.Understanding and applying the power rule helps in quickly finding derivatives and is a crucial skill for solving calculus problems involving polynomials and power functions.