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: Determine the value of \(g'(x)\) where \(g(x) = f(\sin x)\), given that \(f'(0) = 3\) and \(f'(1) = 5\) for the following points: a. \(x = 0\) b. \(x = \frac{\pi}{2}\) c. \(x = \pi\) Answer: a. \(g'(0) = 3\) b. \(g'\left(\frac{\pi}{2}\right) = 0\) c. \(g'(\pi) = -3\)

Step-by-step solution

Find the derivative of g(x)

To find the derivative \(g'(x)\), we use the chain rule for composed functions, which states that if \(g(x) = f(h(x))\), then \(g'(x) = f'(h(x))h'(x)\). In our case, \(h(x) = \sin x\), so we have: \(g'(x) = f'(\sin x)\cdot \sin' x\) Now, since \(f\) is differentiable on \([-2, 2]\), we are given \(f'(0) = 3\) and \(f'(1) = 5\). We also know the derivative of the sine function, \( \sin' x = \cos x\). So, we get: \(g'(x) = f'(\sin x)\cdot \cos x\) Now, we move on to calculating the different expressions in parts a, b, and c.

Calculate \(g'(0)\)

To find \(g'(0)\), we plug in 0 into the expression we obtained for \(g'(x)\): \(g'(0) = f'(\sin 0)\cdot \cos 0\) Since \(\sin 0 = 0\) and \(\cos 0 = 1\), we have: \(g'(0) = f'(0)\cdot 1\) As given, \(f'(0) = 3\), so: \(g'(0) = 3\)

Calculate \(g'\left(\frac{\pi}{2}\right)\)

To find \(g'\left(\frac{\pi}{2}\right)\), we plug in \(\frac{\pi}{2}\) into the expression we obtained for \(g'(x)\): \(g'\left(\frac{\pi}{2}\right) = f'(\sin \frac{\pi}{2})\cdot \cos \frac{\pi}{2}\) As \(\sin \frac{\pi}{2} = 1\) and \(\cos \frac{\pi}{2} = 0\), we have: \(g'\left(\frac{\pi}{2}\right) = f'(1)\cdot 0\) As given, \(f'(1) = 5\), so: \(g'\left(\frac{\pi}{2}\right) = 5 \cdot 0\) \(g'\left(\frac{\pi}{2}\right) = 0\)

Calculate \(g'(\pi)\)

To find \(g'(\pi)\), we plug in \(\pi\) into the expression we obtained for \(g'(x)\): \(g'(\pi) = f'(\sin \pi)\cdot \cos \pi\) As \(\sin \pi = 0\) and \(\cos \pi = -1\), we have: \(g'(\pi) = f'(0)\cdot (-1)\) As given, \(f'(0) = 3\), so: \(g'(\pi) = 3\cdot(-1)\) \(g'(\pi) = -3\) The solutions to the exercise are: a. \(g'(0) = 3\) b. \(g'\left(\frac{\pi}{2}\right) = 0\) c. \(g'(\pi) = -3\)

Key concepts

Chain Rule

The chain rule is an essential concept in calculus that helps us find the derivative of composite functions. A composite function is formed when one function is applied and then another. For instance, if we have two functions, \( f(x) \) and \( h(x) \), and we want to find the derivative of \( f(h(x)) \), we use the chain rule. The formula for the chain rule is:

• If \( g(x) = f(h(x)) \), then \( g'(x) = f'(h(x)) \cdot h'(x) \).

Breaking it down:

• First, find the derivative of the outer function, \( f \), and evaluate it at the inner function \( h(x) \).
• Next, find the derivative of the inner function, \( h(x) \), and then multiply the results together.

The chain rule is used widely because many real-world problems involve such nested or composed functions. It simplifies the process of differentiation, ensuring we get accurate results, such as in the given exercise where \( g(x) = f(\sin x) \). Applying the chain rule allowed us to find the derivative of this composite function easily.

Derivative

A derivative represents the rate at which a function is changing at any given point and is a fundamental tool in calculus. The derivative of a function \( f(x) \) is denoted by \( f'(x) \) or \( \frac{df}{dx} \). It tells us the slope of the tangent line to the curve of the function at any point.

• The derivative gives us vital insights into the function's behavior, such as where it increases or decreases and where it may have maxima or minima.
• For example, in the exercise, knowing \( f'(0) = 3 \) tells us that at \( x = 0 \), the slope of \( f(x) \) is 3, indicating a local influence on \( g(x) = f(\sin x) \).

To compute a derivative, established rules like the power rule, product rule, quotient rule, and the chain rule (discussed earlier) are used. These rules simplify the differentiation of even complex functions so that we can analyze them effectively.

Trigonometric Derivatives

Trigonometric derivatives are derivatives of trigonometric functions like sine, cosine, and others. Knowing these basic derivatives is crucial, as they often appear in various problems across calculus.For the sine and cosine functions, the derivatives are:

• \( \frac{d}{dx}(\sin x) = \cos x \)
• \( \frac{d}{dx}(\cos x) = -\sin x \)

In the given exercise, the derivative of \( \sin x \) is \( \cos x \), which we use to find \( g'(x) \). Each time we perform a derivative involving trigonometric functions, we apply these known derivatives.These results are extended using the chain rule when trigonometric functions are composed with others, as in \( f(\sin x) \). Understanding these derivatives assists in solving a wide range of calculus problems involving periodic or oscillating phenomena.

Differentiable Functions

A function is said to be differentiable at a point if it has a derivative at that point. Differentiability is a key concept indicating that function is smooth at certain points without any breaks, cusps, or vertical tangents.For the function \( f(x) \), if it is differentiable on an interval like \([-2, 2]\), you can find \( f'(x) \) for all points within this interval.

• It implies continuity, meaning without sudden jumps in the values, and the ability to draw a tangent line at any point within that interval.
• For the exercise, \( f \) is differentiable on \([-2, 2] \) means we have the derivatives like \( f'(0) = 3 \) and \( f'(1) = 5 \), allowing us to evaluate \( g'(x) \) accurately especially at critical points like \( 0, \frac{\pi}{2}, \) and \( \pi \).

Differentiable functions are essential in calculus as they model real-world phenomena smoothly, providing insights and predictions based on their rates of change.