Problem 57 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 47-58, express the indicated derivative in terms of the function \(F(x)\). Assume that \(F\) is differentiable. $$ D_{x}\left(F(x) \sin ^{2} F(x)\right) $$

Step-by-Step Solution

Answer

The derivative is \( F'(x) [ \sin^2 F(x) + F(x) \sin(2F(x)) ] \).

Step 1: Identify the Expression

The expression given in the problem is \( F(x) \sin^2 F(x) \). The task is to differentiate this with respect to \( x \).

Step 2: Apply the Product Rule

The expression \( F(x) \sin^2 F(x) \) is a product of two functions: \( F(x) \) and \( \sin^2 F(x) \). According to the product rule, for functions \( u(x) \) and \( v(x) \), the derivative \( D_x(uv) = u'v + uv' \). Here, let \( u = F(x) \) and \( v = \sin^2 F(x) \).

Step 3: Differentiate \( u = F(x) \)

The derivative of \( u = F(x) \) with respect to \( x \) is \( u' = F'(x) \).

Step 4: Differentiate \( v = \sin^2 F(x) \)

The function \( v = \sin^2 F(x) \) can be written as \( (\sin F(x))^2 \). Use the chain rule to differentiate: - First, differentiate \( (\sin F(x))^2 \) as \( 2\sin F(x) \cdot \cos F(x) \). - Then, multiply by the derivative of \( \sin F(x) \), which is \( \cos F(x) \cdot F'(x) \). So, \( v' = 2\sin F(x)\cos F(x) \cdot F'(x) \).

Step 5: Combine Derivatives Using Product Rule

Substitute the derivatives \( u' \) and \( v' \) back into the product rule formula: \( \frac{d}{dx} (F(x)\sin^2 F(x)) = F'(x) \cdot \sin^2 F(x) + F(x) \cdot (2\sin F(x) \cos F(x) \cdot F'(x)) \).

Step 6: Simplify the Expression

Simplify the expression: - Notice that \( 2\sin F(x)\cos F(x) \) is \( \sin(2F(x)) \). - Therefore, the expression simplifies to \( F'(x) \sin^2 F(x) + F(x) \sin(2F(x)) F'(x) \).- Further simplify to \( F'(x) [ \sin^2 F(x) + F(x) \sin(2F(x)) ] \).

Key Concepts

Product Rule

Differentiating products of functions in calculus is handled by the Product Rule. When you need to find the derivative of a product of two functions, like in our problem with \( F(x) \) and \( \sin^2 F(x) \), the Product Rule is your go-to tool.
To apply the rule, you use the formula:

• For two functions \( u(x) \) and \( v(x) \), their derivative is given by \( u'v + uv' \).
• This means you first differentiate \( u \), then multiply by \( v \), plus \( u \) times the derivative of \( v \).

In our case, let \( u = F(x) \) and \( v = \sin^2 F(x) \). It results in:\
\[D_x (F(x) \sin^2 F(x)) = F'(x) \sin^2 F(x) + F(x) v'\]where \( v' \) is found using another rule we will discuss next.

Chain Rule

In calculus, the Chain Rule is a vital tool for differentiating composite functions. A composite function is like a nested function – one function inside another.
Using the Chain Rule allows us to handle functions like \( \sin^2 F(x) \), which can be expressed as \( (\sin F(x))^2 \). This rule tells us:

• To differentiate \( (g(h(x))) \), you first differentiate the outer function \( g \) while keeping the inner part \( h(x) \) unchanged, then multiply by the derivative of the inner function \( h(x) \).

For our problem, we treat \( v = \sin^2 F(x) \) as \( (\sin F(x))^2 \), and use:

• The derivative of \( x^2 \) is \( 2x \), thus for our function, \( 2\sin F(x) \cdot \cos F(x) \cdot F'(x) \).

It gives us the derivative \( v' = 2\sin F(x)\cos F(x) \cdot F'(x) \), which fits back into our Product Rule formula.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, often appear in calculus problems, and it's crucial to know how to differentiate them.
In this exercise, we work with \( \sin F(x) \) and \( \cos F(x) \). These functions have straightforward derivatives:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \(-\sin x \).

They are essential when using the Chain Rule or Product Rule in tandem with trigonometric expressions.
For example, in our case:

• \( \sin F(x) \) differed at part of the Chain Rule derivation becomes \( \cos F(x) \cdot F'(x) \).

These trigonometric derivatives hence contribute to constructing parts of the overall derivative.

Derivative Simplification

Once you have applied the necessary rules to find a derivative, simplification helps make the expression cleaner and more understandable.
For the problem at hand, the derivative ended up as:

• \( F'(x) \sin^2 F(x) + F(x) \cdot (2\sin F(x) \cos F(x) \cdot F'(x)) \)

A key observation can simplify this expression.
Note that the expression \( 2\sin F(x)\cos F(x) \) is equivalent to the sine double-angle formula \( \sin(2F(x)) \).
This recognition reduces the expression to:
\[F'(x) [\sin^2 F(x) + F(x) \sin(2F(x)) ]\]Understanding and using trigonometric identities efficiently can significantly simplify complex derivative results.