Problem 52
Question
Find the derivative with respect to the independent variable. $$ f(x)=\left(2 x^{3}-x\right) \cos ^{2} x $$
Step-by-Step Solution
Verified Answer
The derivative is \( f'(x) = (6x^2 - 1)\cos^2 x - (2x^3 - x)\sin 2x \).
1Step 1: Apply the Product Rule
The function given is a product of two functions: \( g(x) = 2x^3 - x \) and \( h(x) = \cos^2 x \). According to the product rule, if \( f(x) = u(x) v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \). So, we need to find the derivatives \( g'(x) \) and \( h'(x) \).
2Step 2: Differentiate \( g(x) \)
Find the derivative of \( g(x) = 2x^3 - x \). Differentiate term by term:\( g'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(x) = 6x^2 - 1 \).
3Step 3: Simplify \( h(x) \) and Differentiate
Use the trigonometric identity \( \cos^2 x = (\cos x)(\cos x) \) and apply the chain rule to differentiate:\( h'(x) = \frac{d}{dx}(\cos^2 x) = 2\cos x (-\sin x) = -2\cos x \sin x \).This simplifies to \( -\sin(2x) \), recalling the double angle identity: \( \sin(2x) = 2 \sin x \cos x \).
4Step 4: Substitute and Simplify
Substitute \( g(x) \), \( g'(x) \), \( h(x) \), and \( h'(x) \) into the product rule formula:\[ f'(x) = (6x^2 - 1)(\cos^2 x) + (2x^3 - x)(-\sin 2x) \]This simplification involves distributing and combining like terms.
Key Concepts
Product RuleChain RuleTrigonometric Identities
Product Rule
The product rule is an essential differentiation formula used when dealing with functions that are multiplied together. If you have two functions, say \( u(x) \) and \( v(x) \), the derivative of their product is not simply the product of their derivatives. Instead, the product rule states:
For our exercise, the function \( f(x) = (2x^3 - x) \cos^2 x \) involves two components: \( g(x) = 2x^3 - x \) and \( h(x) = \cos^2 x \). We first differentiate each part separately and then apply the rule to find the derivative of the entire expression.
The product rule helps ensure that all parts of the function's interaction are considered, leading to the correct derivative.
- If \( f(x) = u(x) v(x) \), then \( f'(x) = u'(x) v(x) + u(x) v'(x) \).
For our exercise, the function \( f(x) = (2x^3 - x) \cos^2 x \) involves two components: \( g(x) = 2x^3 - x \) and \( h(x) = \cos^2 x \). We first differentiate each part separately and then apply the rule to find the derivative of the entire expression.
The product rule helps ensure that all parts of the function's interaction are considered, leading to the correct derivative.
Chain Rule
The chain rule is incredibly useful when differentiating composite functions, where one function is nested inside another. The general principle of the chain rule can be summarized as:
In the case of our exercise, the function part \( h(x) = \cos^2 x \) is composite since it involves the square of a cosine function. To differentiate this:
- If a function can be represented as \( f(g(x)) \), then its derivative \( f'(g(x)) \) is: \( f'(g(x)) \cdot g'(x) \).
In the case of our exercise, the function part \( h(x) = \cos^2 x \) is composite since it involves the square of a cosine function. To differentiate this:
- Firstly, view it as \( ( ext{cos}(x))^2 \).
- Secondly, differentiate the outside power \( 2( ext{cos}(x)) \), resulting in \( 2 \cdot \text{cos}(x) \). Then multiply by the derivative of \( \cos(x) \), which is \(-\sin(x)\).
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold for all values of the variable. They can simplify the differentiation process, uncovering simpler forms, and connecting different trigonometric expressions.
In our exercise, we encounter an identity during the differentiation of \( \cos^2 x \) using the chain rule. When we differentiated this, we achieved \( -2\cos(x)\sin(x) \). Using a well-known identity, we simplify this expression to:
Understanding and using these identities efficiently help to make complex derivatives simpler and more manageable. They are some of the most powerful tools in the calculus toolbox, offering shortcuts and clarity when dealing with trigonometric functions and their derivatives.
In our exercise, we encounter an identity during the differentiation of \( \cos^2 x \) using the chain rule. When we differentiated this, we achieved \( -2\cos(x)\sin(x) \). Using a well-known identity, we simplify this expression to:
- The double angle identity: \( \sin(2x) = 2\sin(x)\cos(x) \).
Understanding and using these identities efficiently help to make complex derivatives simpler and more manageable. They are some of the most powerful tools in the calculus toolbox, offering shortcuts and clarity when dealing with trigonometric functions and their derivatives.
Other exercises in this chapter
Problem 51
Differentiate with respect to the independent variable. \(f(x)=\frac{3 x^{2}-2 x+1}{2 x+1}\)
View solution Problem 51
Find the normal line, in standard form, to \(y=f(x)\) at the indicated point. $$ y=\sqrt{3} x^{4}-2 \sqrt{3} x^{2}, \text { at } x=-\sqrt{3} $$
View solution Problem 52
Differentiate the functions with respect to the independent variable. \(g(r)=4^{r^{1 / 4}}\)
View solution Problem 52
Differentiate with respect to the independent variable. \(f(x)=\frac{x^{4}+2 x-1}{5 x^{2}-2 x+1}\)
View solution