Question

Express the indicated derivative in terms of the function \(F(x) .\) Assume that \(F\) is differentiable. $$ D_{x}\left(F(x) \sin ^{2} F(x)\right) $$

Short answer

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

Step-by-step solution

Apply the Product Rule

The derivative of a product of two functions, say \(u(x)\) and \(v(x)\), is given by the product rule: \[D_x[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\].In this case, we have \(u(x) = F(x)\) and \(v(x) = \sin^2(F(x))\). First, compute the derivative of each function independently.

Differentiate \(F(x)\)

Since \(u(x) = F(x)\), we have \(u'(x) = F'(x)\) because the derivative of a function with respect to itself is simply the derivative of that function.

Differentiate \(\sin^2(F(x))\)

For \(v(x) = \sin^2(F(x))\), apply the chain rule: \[ \frac{d}{dx}[ ext{sin}^2(F(x))] = 2\sin(F(x))\cdot\cos(F(x))\cdot F'(x) \].Here, we first differentiate \(\sin^2(u)\) with respect to \(u\), resulting in \(2\sin(u)\cos(u)\), and then multiply by \(F'(x)\) as per the chain rule, where \(u = F(x)\).

Substitute Back into Product Rule

Substitute the derivatives from steps 2 and 3 into the product rule formula: \[D_x[F(x) \sin^2(F(x))] = F'(x) \cdot \sin^2(F(x)) + F(x) \cdot (2\sin(F(x))\cos(F(x))F'(x) )\].

Simplify the Expression

Factor out \(F'(x)\) from the expression:\[D_x[F(x) \sin^2(F(x))] = F'(x)(\sin^2(F(x)) + 2F(x) \sin(F(x))\cos(F(x))).\] Recognize that \(2\sin(F(x))\cos(F(x)) = \sin(2F(x))\), so further simplify:\[D_x[F(x) \sin^2(F(x))] = F'(x)(\sin^2(F(x)) + F(x)\sin(2F(x))).\]

Key concepts

Product Rule

In calculus, the product rule is essential when finding the derivative of a product of two functions. The product rule states that if you have two differentiable functions, say \( u(x) \) and \( v(x) \), the derivative of their product \( u(x)v(x) \) is:

• \( u'(x)v(x) + u(x)v'(x) \)

This rule helps you differentiate products efficiently by first differentiating one function while holding the other constant and then switching.

Let's see an example. For our exercise, we set \( u(x) = F(x) \) and \( v(x) = \sin^2(F(x)) \). Each part's derivative is calculated, and these are substituted into the product rule formula. This is how we efficiently handle products of functions in calculus differentiation.

Using the product rule early in a calculation can simplify differentiation, saving time and reducing complex manual calculations. It's like breaking a intricate task into two smaller, more manageable chores.

Calculus Differentiation

Differentiation is a central concept in calculus, involving the process of finding the rate at which a function changes at any given point.

It is represented by the derivative \( f'(x) \), which provides meaningful insights into the behavior of the function. Differentiation allows you to calculate things like slope, velocity, and other rates of change.

• The derivative of a standard function gives you slope.
• The power of calculus differentiation is its ability to be applied to a wide range of functions and scenarios.
• It's also the fundamental step towards understanding more complex calculus concepts.

In practice, differentiation involves using rules like the product, quotient, and chain rule to compute derivatives accurately.

Always remember that when you're differentiating a function, it's all about finding how the function behaves as its input changes.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are mathematical functions that relate the angles of a triangle to the lengths of its sides.

They are crucial in calculus due to their periodic nature and versatility in modeling wave-like phenomena. When differentiating trigonometric functions, it's important to grasp their derivatives:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• The derivative of \( \tan(x) \) is \( \sec^2(x) \).

In our exercise, we dealt with \( \sin^2(F(x)) \), which required using the chain rule.

This helped us tackle the composite nature of the trigonometric expression correctly. Understanding the behavior of trigonometric functions in calculus can be tricky, but by learning their properties and derivatives, you can solve complex problems involving oscillations, waves, and circular motion.