Question

Suppose \(f\) is differentiable on [-2,2] with \(f^{\prime}(0)=3\) and \(f^{\prime}(1)=5 .\) Let \(g(x)=f(\sin x)\) Evaluate the following expressions. a. \(g^{\prime}(0)\) b. \(g^{\prime}\left(\frac{\pi}{2}\right)\) c. \(g^{\prime}(\pi)\)

Short answer

Question: Given \(g(x) = f(\sin x)\) with known derivatives \(f'(0) = 3\) and \(f'(1) = 5\), find \(g'(0)\), \(g'\left(\frac{\pi}{2}\right)\), and \(g'(\pi)\). Answer: Using the chain rule, we find that \(g'(0) = 3\), \(g'\left(\frac{\pi}{2}\right) = 0\), and \(g'(\pi) = -3\).

Step-by-step solution

Use the chain rule

As \(g(x) = f(\sin x)\), we need to use the chain rule to find its derivative: \(g'(x) = \frac{d}{dx}\left[f(\sin x)\right] = f'(\sin x)\cdot\cos x\). Now, we can find the value of \(g'(0)\).

Evaluate the derivative at \(x=0\)

To find \(g'(0)\), substitute \(x=0\) in the derivative formula: \(g'(0) = f'(\sin 0)\cdot\cos 0 = f'(0)\cdot 1\). Since \(f'(0) = 3\), we have: \(g'(0) = 3\). b. Find \(g'\left(\frac{\pi}{2}\right)\)

Evaluate the derivative at \(x=\frac{\pi}{2}\)

To find \(g'\left(\frac{\pi}{2}\right)\), substitute \(x=\frac{\pi}{2}\) in the derivative formula: \(g'\left(\frac{\pi}{2}\right) = f'(\sin \frac{\pi}{2})\cdot\cos \frac{\pi}{2} = f'(1)\cdot 0\). Since \(f'(1) = 5\), we have: \(g'\left(\frac{\pi}{2}\right) = 0\). c. Find \(g'(\pi)\)

Evaluate the derivative at \(x=\pi\)

To find \(g'(\pi)\), substitute \(x=\pi\) in the derivative formula: \(g'(\pi) = f'(\sin \pi)\cdot\cos \pi = f'(0)\cdot (-1)\). Since \(f'(0)=3\), we have: \(g'(\pi) = -3\).

Key concepts

Understanding Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes at any given point. Essentially, when we differentiate a function, we are calculating its derivative, which tells us how the function's value changes with respect to changes in its input variables.

To differentiate a function, we apply certain rules and techniques that have been developed over time, with the objective being to systematically find the derivative. Differentiation rules such as the product rule, quotient rule, and most importantly for this example, the chain rule, allow us to differentiate complex functions systematically. Differentiation is not just limited to finding the slope of a tangent; it's a powerful tool used in various fields like physics, engineering, and economics, allowing us to calculate accelerations, optimize solutions, and model real-world scenarios effectively.

Exploring Derivatives

The derivative of a function at a particular point gives us the slope of the tangent line to the function at that point. It essentially measures the function's sensitivity to changes in the independent variable.

When we say a function is differentiable, it's smooth and continuous, allowing us to find derivatives using calculus. For example, in our exercise, the given function is differentiable over the interval [-2,2], which allows us to effectively use the chain rule to find its derivative. Differentiability ensures that we can precisely calculate the instantaneous rate of change.

It's important to note some key features about derivatives:

• A derivative can be considered as the slope of a curve at a specific point.
• It helps predict how one quantity changes in response to another.
• Higher-order derivatives represent the rate of change of the rate of change.

Derivatives are central to calculus and have numerous applications in evaluating and tackling complex problems involving change.

Understanding Composite Functions and the Chain Rule

When dealing with composite functions, which are functions made up of other functions, differentiation becomes an intricate process. The chain rule specifically helps us address this complexity by providing a formula to differentiate such composite functions.

In the context of the exercise, we have a composite function given by \(g(x) = f(\sin x)\). Here, \(f\) and \(\sin x\) are the inner and outer functions, respectively. The chain rule states that to find the derivative of a composite function, you must:

• Differentiate the outer function, evaluated at the inner function.
• Multiply that by the derivative of the inner function.

Using the formula \( g'(x) = f'(\sin x) \cdot \cos x \), we apply the derivatives of both inner and outer functions.

This rule is vital for working with nested functions, such as \(f(\sin x)\), where one function is wrapped inside another. Mastering the chain rule is essential for students learning calculus, as it opens doors to effectively differentiating a wide variety of complex functions in mathematics.