Problem 15
Question
$$ \underline{\phantom{xxx}} \text { find } D_{x} y . $$ $$ y=x^{2} \cos x $$
Step-by-Step Solution
Verified Answer
\( D_x y = 2x \cos x - x^2 \sin x \)
1Step 1: Identify the Rule to Use
The expression to differentiate is a product of two functions: \( y = x^2 \) and \( y = \cos x \). Thus, we use the product rule for differentiation, which states \( D_x (u imes v) = u'v + uv' \).
2Step 2: Differentiate Each Function Separately
First, differentiate \( u = x^2 \) to get \( u' = 2x \). Next, differentiate \( v = \cos x \) to get \( v' = -\sin x \).
3Step 3: Apply the Product Rule
Substitute \( u = x^2 \), \( u' = 2x \), \( v = \cos x \), and \( v' = -\sin x \) into the product rule formula: \( u'v + uv' = (2x)(\cos x) + (x^2)(-\sin x) \).
4Step 4: Simplify the Expression
Perform the multiplications and simplify: \( 2x \cos x - x^2 \sin x \). Therefore, \( D_x y = 2x \cos x - x^2 \sin x \).
Key Concepts
Differentiation TechniquesTrigonometric FunctionsDerivatives Computation
Differentiation Techniques
Differentiation is a fundamental concept in calculus, capturing the idea of how a function changes as its input changes. Among the various methods used in differentiation, one crucial technique is the product rule. This rule is essential when dealing with expressions that are products of two functions.
To recall, the product rule formula is given by \( D_x (u \times v) = u'v + uv' \). Here, \( u \) and \( v \) are the functions being multiplied, while \( u' \) and \( v' \) are their respective derivatives. Using this approach ensures that the derivative takes into account how both functions are changing with respect to the variable.
Let's take a simple example:
To recall, the product rule formula is given by \( D_x (u \times v) = u'v + uv' \). Here, \( u \) and \( v \) are the functions being multiplied, while \( u' \) and \( v' \) are their respective derivatives. Using this approach ensures that the derivative takes into account how both functions are changing with respect to the variable.
Let's take a simple example:
- If \( f(x) = x^3 \sin x \), first identify \( u = x^3 \) and \( v = \sin x \).
- Differentiate them to obtain \( u' = 3x^2 \) and \( v' = \cos x \).
- Apply the product rule resulting in: \( D_x f = 3x^2 \sin x + x^3 \cos x \).
Trigonometric Functions
Trigonometric functions like sine, cosine, and tangent are vital in both mathematics and applied sciences. These functions describe the ratios of sides in right triangles and have periodic properties, making them relevant in analyzing wave patterns, circular motions, and more.
The cosine function, denoted as \( \cos x \), is the focus of this discussion. This function oscillates between -1 and 1, repeating every \( 2\pi \) radians. In differentiation, the derivative of the cosine function is important to know and easily remembered as \( -\sin x \).
Understanding trigonometric derivatives enables us to address problems that involve periodic or oscillatory behaviors. Here's a quick guide to trigonometric derivatives:
The cosine function, denoted as \( \cos x \), is the focus of this discussion. This function oscillates between -1 and 1, repeating every \( 2\pi \) radians. In differentiation, the derivative of the cosine function is important to know and easily remembered as \( -\sin x \).
Understanding trigonometric derivatives enables us to address problems that involve periodic or oscillatory behaviors. Here's a quick guide to trigonometric derivatives:
- The derivative of \( \sin x \) is \( \cos x \).
- The derivative of \( \cos x \) is \( -\sin x \).
- The derivative of \( \tan x \) is \( \sec^2 x \).
Derivatives Computation
Derivatives capture the concept of rate of change and slope of a curve at any given point. Computing derivatives involves applying rules and understanding the fundamental behavior of the function under consideration.
When solving the problem \( y = x^2 \cos x \), we calculated the derivative using both the product rule and knowledge of basic derivatives. Here's a breakdown of the computation:
When solving the problem \( y = x^2 \cos x \), we calculated the derivative using both the product rule and knowledge of basic derivatives. Here's a breakdown of the computation:
- Firstly, identify the components: \( u = x^2 \) and \( v = \cos x \).
- Compute their derivatives separately: \( u' = 2x \) and \( v' = -\sin x \).
- Apply the product rule: \( D_x y = (2x)(\cos x) + (x^2)(-\sin x) \).
- Simplify the result to \( 2x \cos x - x^2 \sin x \).
Other exercises in this chapter
Problem 15
Find \(f^{\prime \prime}(2)\). $$ f(s)=s\left(1-s^{2}\right)^{3} $$
View solution Problem 15
Find \(D_{x} y\). $$ y=\cos \left(\frac{3 x^{2}}{x+2}\right) $$
View solution Problem 15
Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). $$ F(x)=\frac{6}{x^{2}+1} $$
View solution Problem 15
Find \(D_{x} y\) using the rules of this section. $$ y=\pi x^{7}-2 x^{5}-5 x^{-2} $$
View solution