Question

Computing the derivative of \(f(x)=x^{2} e^{x}\) a. Use the definition of the derivative to show that \(\frac{d}{d x}\left(x^{2} e^{x}\right)=e^{x} \cdot \lim _{h \rightarrow 0} \frac{\left(x^{2}+2 x h+h^{2}\right) e^{h}-x^{2}}{h}\) b. Manipulate the limit in part (a) to arrive at \(f^{\prime}(x)=e^{x}\left(x^{2}+2 x\right) \cdot(\text { Hint}:\) Use the fact that \(\left.\lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=1 .\right)\)

Short answer

Question: Find the derivative of the function \(f(x)=x^{2}e^{x}\) using the definition of the derivative and apply the hint: \(\lim_{h\rightarrow 0} \frac{e^h-1}{h}=1\). Answer: The derivative of \(f(x)=x^{2}e^{x}\) is \(f^{\prime}(x)=e^{x}(x^2+2x)\).

Step-by-step solution

Apply the definition of the derivative

First, recall the definition of the derivative: \(f^{\prime}(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}\). Now, substitute \(f(x)=x^{2}e^{x}\) into the definition: \(f^{\prime}(x)=\lim_{h\rightarrow 0} \frac{(x+h)^{2}e^{x+h}-x^{2}e^{x}}{h}\).

Expand the numerator

In order to simplify the expression, we need to expand the numerator: \(\begin{aligned} (x+h)^{2}e^{x+h} &= \left(x^{2}+2xh+h^{2}\right)e^{x}e^{h} \\ x^{2}e^{x} &= x^{2}e^{x} \end{aligned}\) Substitute these expressions back into the derivative: \(f^{\prime}(x)=\lim_{h\rightarrow 0} \frac{e^{x}e^{h}\left(x^{2}+2xh+h^{2}\right)-x^{2}e^{x}}{h}\).

Factor out \(e^x\)

Factor \(e^x\) from the numerator: \(f^{\prime}(x)=\lim_{h\rightarrow 0} \frac{e^{x}\left(\left(x^{2}+2xh+h^{2}\right)e^{h}-x^{2}\right)}{h}\). Notice that we have the expression from part (a) as required.

Use the given hint

Now, we will use the hint from the problem, which states that \(\lim_{h\rightarrow 0} \frac{e^h-1}{h}=1\). First, rewrite the numerator: \(\left(x^{2}+2xh+h^{2}\right)e^{h}-x^{2}=\left(x^{2}+2xh+h^{2}\right)(e^{h}-1)+2xh+h^{2}\) Now, divide by \(h\) and apply the limit: \(\begin{aligned} \lim_{h\rightarrow 0}\frac{\left(x^{2}+2xh+h^{2}\right)(e^{h}-1)+2xh+h^{2}}{h} &= \lim_{h\rightarrow 0}\left(\frac{(x^{2}+2xh+h^{2})(e^{h}-1)}{h}+2x+ h\right) \\ &= \lim_{h\rightarrow 0}((x^{2}+2xh+h^{2})\cdot\frac{e^{h}-1}{h})+2x \\ &= (x^{2}+2x)(1)+2x \\ &= x^{2}+2x \end{aligned}\)

Write the final answer

Finally, substitute the simplified limit back into the expression we found in Step 3: \(f^{\prime}(x)=e^{x}(x^{2}+2x)\). Thus, the derivative of \(f(x)=x^{2}e^{x}\) is \(f^{\prime}(x)=e^{x}(x^{2}+2x)\).

Key concepts

Definition of Derivative

Understanding the definition of the derivative is pivotal for calculus students. In simple terms, the derivative is a measure of how a function changes as its input changes. Formally, it is defined as the limit of the average rate of change of the function over an interval as the interval becomes infinitesimally small.

For a function \(f(x)\), its derivative at \(x\), denoted \(f'(x)\) or \(\frac{d}{dx}f(x)\), is given by the limit \(f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}\). This approach, known as the limit definition of a derivative, involves calculating the instantaneous rate of change at a single point by considering an infinitesimal increment, \(h\), to the function's input.

Particularly in our exercise, applying this definition to \(f(x) = x^2e^x\) requires a careful expansion and simplification of the function's value at \(x+h\) and at \(x\), followed by examining the behavior of the resulting expression as \(h\) approaches zero.

Limit Manipulation

The concept of limit manipulation involves algebraically transforming the expression under a limit to make the evaluation of the limit straightforward or even possible. In the context of differentiating functions, it is common to encounter limits that initially appear complex.

In our example, a direct application of the derivative definition leads to an expression that, without manipulation, is not immediately clear how to evaluate as \(h\) approaches zero. Yet, by expanding the terms and then systematically simplifying, we uncover a form that reveals the core principle that \(\lim_{h\rightarrow 0} \frac{e^h-1}{h} = 1\). This manipulation allows us to isolate and eliminate the factor of \(h\) in the denominator, using known limit properties and ultimately simplifying the overall expression.

Product Rule

The product rule is an essential differentiation tool when dealing with the derivatives of products of two functions. It states that for two differentiable functions \(u(x)\) and \(v(x)\), the derivative of their product \(u(x)\cdot v(x)\) is given by \(u'(x)\cdot v(x) + u(x)\cdot v'(x)\).

However, for exponential functions like ours, \(x^2e^x\), the product rule is implied in the process of limit manipulation. When we expanded the numerator, the product of \(x^2\) and \(e^x\) was technically handled by considering the limit definition with product-like terms. Thus, while not explicitly stated in the steps, understanding the product rule provides insight into why we can differentiate functions that are products of two separate functions, such as polynomials and exponentials.

Exponential Differentiation

The process of exponential differentiation is straightforward when working with the common base \(e\), the mathematical constant approximately equal to 2.71828. The derivative of the exponential function \(e^x\) is unique since it is the same as the function itself, \(e^x\).

Exponential differentiation becomes more interesting when the exponential function is multiplied by another function, as in our exercise with \(f(x) = x^2e^x\). Here we not only apply the exponential rule, where we understand that the derivative of \(e^x\) remains \(e^x\), but also combine it with the limit principles and possibly the product rule to find the derivative of the entire function. As shown in the solution, through limit manipulation and simplification, we were able to effectively compute the derivative of this combined function, landing on the result \(f'(x) = e^x(x^2 + 2x)\).