Problem 16
Question
Differentiate the functions and find the slope of the tangent line at the given value of the independent variable. $$y=\frac{x+3}{1-x}, x=-2$$
Step-by-Step Solution
Verified Answer
The slope of the tangent line at \( x = -2 \) is \( \frac{4}{9} \).
1Step 1: Identify the function and apply the quotient rule
Given the function \( y = \frac{x+3}{1-x} \), we first identify it consists of two functions: the numerator \( u(x) = x+3 \) and the denominator \( v(x) = 1-x \). To differentiate this function, we apply the quotient rule, which is \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \).
2Step 2: Differentiate numerator and denominator
Next, we differentiate the numerator and denominator functions. For the numerator, \( u(x) = x+3 \), we find \( u'(x) = 1 \). For the denominator, \( v(x) = 1-x \), we find \( v'(x) = -1 \).
3Step 3: Apply the quotient rule
Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the quotient rule formula: \[ \frac{d}{dx} \left( \frac{x+3}{1-x} \right) = \frac{(1-x) \cdot 1 - (x+3) \cdot (-1)}{(1-x)^2} \]. Simplify the expression: \( \frac{1-x + x + 3}{(1-x)^2} = \frac{4}{(1-x)^2} \).
4Step 4: Evaluate the derivative at x = -2
To find the slope of the tangent line at \( x = -2 \), substitute \( x = -2 \) into the derivative: \[ \frac{4}{(1-(-2))^2} = \frac{4}{3^2} = \frac{4}{9} \].
5Step 5: State the slope of the tangent line
The slope of the tangent line to the function \( y = \frac{x+3}{1-x} \) at the point where \( x = -2 \) is \( \frac{4}{9} \).
Key Concepts
DifferentiationTangent Line SlopeCalculus Problem Solving
Differentiation
Differentiation is a foundational concept in calculus, which allows us to find the rate at which a function changes at any given point. By determining the derivative of a function, we essentially compute the slope of the tangent to the curve represented by the function at a particular point. Differentiation helps us understand how small changes in the independent variable affect the dependent variable.
In the given exercise, we have the function \( y = \frac{x+3}{1-x} \). To differentiate it, we can use the quotient rule, a specific rule developed to differentiate expressions that are fractions of functions. The quotient rule is expressed as \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \), where \( u(x) \) and \( v(x) \) are differentiable functions of \( x \).
This technique simplifies the process of differentiation for complex functions involving division. In this context, the differentiation results in a derivative function that reveals how the original function behaves at every point along its curve.
In the given exercise, we have the function \( y = \frac{x+3}{1-x} \). To differentiate it, we can use the quotient rule, a specific rule developed to differentiate expressions that are fractions of functions. The quotient rule is expressed as \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \), where \( u(x) \) and \( v(x) \) are differentiable functions of \( x \).
This technique simplifies the process of differentiation for complex functions involving division. In this context, the differentiation results in a derivative function that reveals how the original function behaves at every point along its curve.
Tangent Line Slope
The slope of a tangent line is crucial in calculus because it represents the rate of change of a function at a specific point. It is essentially the gradient of the function at that specific location and is determined by the derivative of the function at that point.
In the exercise, once the derivative \( \frac{d}{dx} \left( \frac{x+3}{1-x} \right) = \frac{4}{(1-x)^2} \) is known, we then find the slope of the tangent line by evaluating the derivative at \( x = -2 \).
In the exercise, once the derivative \( \frac{d}{dx} \left( \frac{x+3}{1-x} \right) = \frac{4}{(1-x)^2} \) is known, we then find the slope of the tangent line by evaluating the derivative at \( x = -2 \).
- Substitute \( x = -2 \) into the derivative: \( \frac{4}{(1-(-2))^2} \).
- This yields a slope of \( \frac{4}{9} \).
Calculus Problem Solving
Solving calculus problems often involves several structured steps, allowing you to tackle complex functions systematically. In our exercise, we follow several key problem-solving steps to ensure accuracy.
Firstly, identifying the components of the function correctly sets the stage for applying rules like the quotient rule. This step involves recognizing the numerator \( u(x) = x+3 \) and the denominator \( v(x) = 1-x \).
Next is to compute the derivatives of these components separately. This means differentiating \( u(x) \) to get \( u'(x) \) and differentiating \( v(x) \) to get \( v'(x) \). With these derivatives at hand, we then utilize the quotient rule to find the derivative of the entire function.
Firstly, identifying the components of the function correctly sets the stage for applying rules like the quotient rule. This step involves recognizing the numerator \( u(x) = x+3 \) and the denominator \( v(x) = 1-x \).
Next is to compute the derivatives of these components separately. This means differentiating \( u(x) \) to get \( u'(x) \) and differentiating \( v(x) \) to get \( v'(x) \). With these derivatives at hand, we then utilize the quotient rule to find the derivative of the entire function.
- Our solution requires correctly simplifying the resulting expression.
- Finally, evaluate the derivative at the designated value of \( x \) for the final result, which helps provide specific insights into the function’s behavior at that point.
Other exercises in this chapter
Problem 16
Find \(d y / d x\). $$y=x^{2} \cos x-2 x \sin x-2 \cos x$$
View solution Problem 16
Find \(y^{\prime}\) (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate. $$y=\left(1+x^{2}\ri
View solution Problem 16
Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$h(t)=t^{3}+3 t, \quad\quad(1,4)$$
View solution Problem 17
Find the limits. (If in doubt, look at the function's graph.) $$\lim _{x \rightarrow \infty} \sec ^{-1} x$$
View solution