/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Differentiate the given expressi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Differentiate the given expression with respect to \(x\). \(x e^{x}\)

Short Answer

Expert verified
The derivative of \(x e^{x}\) is \(e^{x}(1 + x)\).

Step by step solution

01

Identify the Function

The given expression is the product of two functions of x: one being the linear polynomial function \(x\) and the other the exponential function \(e^x\). Thus, we recognize that to differentiate this product, we'll need to use the product rule.
02

Recall the Product Rule

The product rule states that if you have a function \(u(x)\) and another function \(v(x)\), and you want to differentiate their product \(u(x)v(x)\), the derivative is given by: \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\).
03

Differentiate Each Component

Identify the components: let \(u(x) = x\) and \(v(x) = e^{x}\). Now, differentiate each component: the derivative of \(u(x) = x\) is \(u'(x) = 1\), and the derivative of \(v(x) = e^{x}\) is \(v'(x) = e^{x}\).
04

Apply the Product Rule

Substitute the derivatives into the product rule formula: \(\frac{d}{dx}[x e^{x}] = (1)(e^{x}) + (x)(e^{x}) = e^{x} + x e^{x}\).
05

Simplify the Expression

In this expression, both terms share a common factor \(e^x\). Therefore, we can factor \(e^x\) out: \(e^x (1 + x)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of Exponential Functions
Exponential functions form a crucial part of calculus and are often difficult for students to grasp initially. At the core, an exponential function is unique because its rate of change is proportional to its current value. This property makes the derivative of the exponential function relatively easy.

For any exponential function of the form \[f(x) = a^{x}, \]its derivative is given by:\[\text{Derivative: } f'(x) = a^{x} \ln(a),\]where \(a\) is a constant. In the special case where \(a = e\), the natural exponential function, the derivative simplifies further, as\[f(x) = e^{x} \quad \text{and hence} \quad f'(x) = e^{x},\]because the natural log of \(e\), \(\ln(e)\), is 1.

This simplicity is inherent only to the natural exponential base \(e\), making it a favorite in both mathematical theory and application.
Differentiation Techniques
Differentiation techniques are essential tools in calculus for finding how a function changes. Different techniques are used based on the type of function you are dealing with to simplify the process and avoid errors.

**Basic Differentiation Rules**
  • Constant Rule: The derivative of a constant is 0.
  • Power Rule: The derivative of \(x^n\) is \(nx^{n-1}\).
  • Sum Rule: The derivative of a sum of functions is the sum of their derivatives.
  • Product Rule: Given functions \(u(x)\) and \(v(x)\), their product derivative is \(u'(x)v(x) + u(x)v'(x)\).
  • Quotient Rule: For two functions, \(u(x)\) and \(v(x)\), the derivative of their division \(\frac{u(x)}{v(x)}\) involves \(\frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\).
The exercise we have uses the **Product Rule**, which is crucial when differentiating a product of functions. Recognizing which differentiation technique to apply is a much-needed skill for solving more complex calculus problems efficiently.
Calculus Problem Solving
Tackling calculus problems can initially seem daunting, but mastering a problem-solving approach makes it manageable and eventually intuitive.

**Approach to Solving Calculus Problems**
  • **Understand the Problem**: Identify what functions you're dealing with and what needs differentiation.
  • **Choose the Right Technique**: Based on the problem format, decide whether to use the power, product, or another differentiation rule.
  • **Work It Out**: Carefully apply the chosen rule. Mistakes often occur in this step, so writing out each piece can help maintain accuracy.
  • **Simplify the Expression**: After applying the differentiation, bring it to the simplest form possible. This helps in understanding and using the solution effectively.
  • **Review Your Solution**: Re-evaluate each step to ensure accuracy. This helps cement understanding and builds confidence.
Developing these methods will aid not only in solving specific problems but also in building a strong foundation in calculus for more advanced studies or applications.

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Most popular questions from this chapter

\- A real-valued function \(f\) of a real variable \(x\) is said to be algebraic if there is a polynomial \(p(u, v)\) with integer coefficients such that \(p(x, f(x))=0\) for all \(x .\) For example, \(f(x)=\) \(2 x+\sqrt{x^{2}+1}\) is algebraic because $$ \begin{aligned} p(x, f(x))=& 3 x^{2}+\left(2 x+\sqrt{x^{2}+1}\right)^{2} \\ &-4 x\left(2 x+\sqrt{x^{2}+1}\right)-1 \\ \equiv & 0 \end{aligned} $$ for \(p(u, v)=3 u^{2}+v^{2}-4 u v-1 .\) A function that is not algebraic is said to be transcendental. Find a polynomial that shows that the given expression is algebraic. $$ 2+\sqrt{1+x / x} $$

Differentiate the given expression with respect to \(x\). $$ \sinh ^{-1}(5 x) $$

Differentiate the given expression with respect to \(x\). $$ \operatorname{coth}\left(x^{2}\right) $$

Calculate the linearization \(L(x)=f(c)+\) \(f^{\prime}(c), \cdot(x-c)\) for the given function \(f\) at the given value \(c\) $$ f(x)=\cos ^{2}(x), c=\pi / 4 $$

A function \(f,\) a point \(c,\) an increment \(\Delta x,\) and a positive integer \(n\) are given. Use the method of increments to estimate \(f(c+\Delta x)\). Then let \(h=\Delta x\) / \(N\). Use the method of increments to obtain an estimate \(y_{1}\) of \(f(c+h) .\) Now, with \(c+h\) as the base point and \(y_{1}\) as the value of \(f(c+h),\) use the method of increments to obtain an estimate \(y_{2}\) of \(f(c+2 h)\). Continue this process until you obtain an estimate \(y_{N}\) of \(f(c+N \cdot h)=f(c+\Delta x) .\) We say that we have taken \(N\) steps to obtain the approximation. The number \(h\) is said to be the step size. Use a calculator or computer to evaluate \(f(c+\Delta x)\) directly. Compare the accuracy of the one step and \(N\) -step approximations. $$ f(x) 1 / \sqrt[3]{x}, c=-8, \Delta x=1, N=4 $$

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