Problem 20 - Thomas Calculus

Question

In Exercises \(17-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=e^{\left(4 \sqrt{x}+x^{2}\right)} $$

Step-by-Step Solution

Answer

The derivative is \(\frac{dy}{dx} = e^{(4\sqrt{x} + x^2)} \cdot (2x^{-1/2} + 2x)\).

Step 1: Identify the Outer Function

The given function is \(y = e^{(4\sqrt{x} + x^2)}\). Here, the outer function is \(e^u\) where \(u = 4\sqrt{x} + x^2\). We will use the chain rule to differentiate with respect to \(x\).

Step 2: Differentiate the Outer Function

Using the chain rule, the derivative of \(e^u\) with respect to \(x\) is \(e^u \cdot \frac{du}{dx}\). Our next step involves finding \(\frac{du}{dx}\).

Step 3: Differentiate the Inner Function

Differentiate \(u = 4\sqrt{x} + x^2\) with respect to \(x\). We break this into two parts: 1. \(4\sqrt{x} = 4x^{1/2}\) becomes \(\frac{d}{dx}(4x^{1/2}) = 4 \cdot \frac{1}{2}x^{-1/2} = 2x^{-1/2}\). 2. \(x^2\) becomes \(\frac{d}{dx}(x^2) = 2x\).

Step 4: Combine Results from Inner Derivative

Sum the derivatives of each part of \(u\):\[\frac{du}{dx} = 2x^{-1/2} + 2x\]

Step 5: Substitute and Simplify

Substitute \(\frac{du}{dx}\) back into the derivative of the outer function:\[\frac{dy}{dx} = e^{(4\sqrt{x} + x^2)} \cdot (2x^{-1/2} + 2x)\]Simplify if necessary, but in this case, it's simplest to leave it in this factored form.

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used for finding the derivatives of composite functions. A composite function is essentially a function within another function.

• For example, consider a function such as \( y = e^{(4\sqrt{x} + x^2)} \), where an "inner" function \( u = 4\sqrt{x} + x^2 \) is plugged into an "outer" function \( e^u \).
• The chain rule states that the derivative of such a composite function is the derivative of the outer function, with respect to the inner function, multiplied by the derivative of the inner function with respect to the variable of interest.

This can be expressed mathematically as:\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.\]In our exercise, this means we first identify and differentiate the outer function, \( e^u \), and then multiply it by the derivative of the inner function \( 4\sqrt{x} + x^2 \).

Exponential Functions

Exponential functions are a class of functions where the variable appears in the exponent. The function \( y = e^x \) is the simplest form of an exponential function, where \( e \) is the base of natural logarithms, approximately \( 2.718 \).

• One of the key properties of exponential functions is that their rate of growth is proportional to their current value.
• This means that the derivative of \( e^u \), where \( u \) is a function of \( x \), is simply \( e^u \cdot \frac{du}{dx} \).

This is powerful because it allows us to handle complex expressions like \( y = e^{(4\sqrt{x} + x^2)} \) by focusing on differentiating the inner function and then multiplying by the unchanged exponential expression. The neat part is that the differentiated exponential function looks very similar to the original, which simplifies work tremendously in calculus.

Derivative Computation

Derivative computation involves taking the limit of the average rate of change of a function as the interval approaches zero. In practical terms, this means finding the slope of the tangent line to the function at a particular point.

• When computing derivatives, we often use rules like the power rule, product rule, and quotient rule, but when dealing with nested functions, the chain rule comes into play.
• In this exercise, after recognizing the function is exponential with an inner function \( u = 4\sqrt{x} + x^2 \), we needed to compute \( \frac{du}{dx} \).

For \( u = 4x^{1/2} + x^2 \), separate the terms to use the power rule:- The derivative of \( 4x^{1/2} \) is \( 2x^{-1/2} \). This results from applying the rule \( \frac{d}{dx}(x^n) = nx^{n-1} \), using \( n = 1/2 \).- The derivative of \( x^2 \) is \( 2x \), using \( n = 2 \).The complete derivative \( \frac{dy}{dx} \) is then formed by substituting these results back into the derivative of the outer function, providing a complete picture of how the function's rate of change behaves.