Problem 59
Question
Given that \(f(0)=1\) and \(f^{\prime}(0)=2\), find \(g^{\prime}(0)\) where \(g(x)=\cos f(x)\)
Step-by-Step Solution
Verified Answer
The value of \(g^{\prime}(0)\) is approximately \(-1.683\).
1Step 1: Identifying the Chain Rule
The function given is composed of an outer function and an inner function. Here, \(g(x) = \cos(f(x))\). To find \(g^{\prime}(x)\), we will need to use the chain rule, which states that if you have a function \(h(x) = u(v(x))\), then \(h^{\prime}(x) = u^{\prime}(v(x)) \cdot v^{\prime}(x)\). In our problem: \(u(v) = \cos(v)\) and \(v(x) = f(x)\).
2Step 2: Differentiating the Outer Function
The derivative of the outer function \(u(v) = \cos(v)\) with respect to \(v\) is \(-\sin(v)\). So, \(u^{\prime}(v) = -\sin(v)\).
3Step 3: Differentiating the Inner Function
The derivative of the inner function \(v(x) = f(x)\) with respect to \(x\) is simply \(f^{\prime}(x)\). Given that \(f^{\prime}(0) = 2\), this is the rate of change of the function at \(x=0\).
4Step 4: Applying the Chain Rule
Apply the chain rule: \[ g^{\prime}(x) = u^{\prime}(v(x)) \cdot v^{\prime}(x) = (-\sin(f(x))) \cdot f^{\prime}(x) \].
5Step 5: Substituting the Known Values
Now, substitute the known values at \(x=0\). You need to find \(g^{\prime}(0)\): \[ g^{\prime}(0) = -\sin(f(0)) \cdot f^{\prime}(0)\]. Given \(f(0) = 1\) and \(f^{\prime}(0) = 2\), substitute these into the equation: \[ g^{\prime}(0) = -\sin(1) \cdot 2\].
6Step 6: Calculating the Derivative
Calculate the numerical value: \(\sin(1)\) is the sine of 1 radian, which is approximately 0.8415. Thus, \[ g^{\prime}(0) = -0.8415 \times 2 = -1.683\].
Key Concepts
DifferentiationComposite FunctionsCalculus Problem Solving
Differentiation
Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes at any given point. Think of differentiation as a tool that helps us understand how one quantity changes in relation to another. For instance, if you have a position-time curve in physics, differentiation helps you find the velocity, which is the rate of change of position with respect to time.
In the original exercise, we are finding the derivative of a composite function. This involves calculating the derivative of functions that are made up of other functions, known as composite functions. By applying the chain rule, we differentiate both the outer and the inner functions of the composite.
In the original exercise, we are finding the derivative of a composite function. This involves calculating the derivative of functions that are made up of other functions, known as composite functions. By applying the chain rule, we differentiate both the outer and the inner functions of the composite.
- The derivative gives us the slope of the tangent line at a particular point on the function's graph.
- It is often written as \(f^{\prime}(x)\) if \(f\) is the function we're differentiating.
Composite Functions
Composite functions occur when one function is applied to the results of another function. Understanding composite functions is key to solving problems that involve more than one layer of mathematical operations, like in our current example where \(g(x) = \cos(f(x))\).
A simple way to see composite functions is to consider them as a machine that takes an input, processes this input with the first machine (function), and then processes the output with the second machine (function). Here, \(f(x)\) is the machine, and \(\cos\) works on the output of that machine.
A simple way to see composite functions is to consider them as a machine that takes an input, processes this input with the first machine (function), and then processes the output with the second machine (function). Here, \(f(x)\) is the machine, and \(\cos\) works on the output of that machine.
- Composite functions are noted as \( g(x) = u(v(x)) \).
- In the exercise, \(u(v) = \cos(v)\) and \(v(x) = f(x)\).
Calculus Problem Solving
Solving calculus problems often involves breaking down complex expressions through rules like the chain rule. The goal is to simplify the differentiation process by identifying the structure of the given functions.
The chain rule is especially useful in differentiation when dealing with composite functions. In our exercise, the problem requires finding the derivative of \(g(x) = \cos(f(x))\), which inherently has an inner \(f(x)\) and an outer function \(\cos\). Understanding the structure allows the solution to be more straightforward and methodical.
The chain rule is especially useful in differentiation when dealing with composite functions. In our exercise, the problem requires finding the derivative of \(g(x) = \cos(f(x))\), which inherently has an inner \(f(x)\) and an outer function \(\cos\). Understanding the structure allows the solution to be more straightforward and methodical.
- Recognize the function layers: identify which is the outer and which is the inner function.
- Calculate their derivatives separately: find both \(u^{\prime}(v)\) and \(v^{\prime}(x)\).
- Apply the chain rule: multiply these derivatives to solve for \(g^{\prime}(x)\).
Other exercises in this chapter
Problem 58
Convince yourself that \(f(x)=\left(x^{x}\right)^{x}\) and \(g(x)=x^{\left(x^{x}\right)}\) are not the same function. Then find \(f^{\prime}(x)\) and \(g^{\prim
View solution Problem 58
In Problems 47-58, express the indicated derivative in terms of the function \(F(x)\). Assume that \(F\) is differentiable. $$ D_{x} \sec ^{3} F(x) $$
View solution Problem 60
Let \(P(a, b)\) be a point on the first quadrant portion of the curve \(y=1 / x\) and let the tangent line at \(P\) intersect the \(x\)-axis at \(A\). Show that
View solution Problem 60
Given that \(F(0)=2\) and \(F^{\prime}(0)=-1\), find \(G^{\prime}(0)\) where \(G(x)=\frac{x}{1+\sec F(2 x)}\)
View solution