Last updated on July 22nd, 2025
We use the derivative of 1/(x+2) to understand how this function changes in response to a slight change in x. Derivatives help us calculate rates of change in various contexts, such as speed in physics or marginal cost in economics. We will now discuss the derivative of 1/(x+2) in detail.
We now understand the derivative of 1/(x+2). It is commonly represented as d/dx (1/(x+2)) or (1/(x+2))', and its value is -1/(x+2)². The function 1/(x+2) has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Rational Function: A function of the form 1/(x+2).
Quotient Rule: Rule for differentiating 1/(x+2) (as it can be seen as 1 divided by a function of x).
Power Rule: Used to differentiate functions of the form xⁿ.
The derivative of 1/(x+2) can be denoted as d/dx (1/(x+2)) or (1/(x+2))'. The formula we use to differentiate 1/(x+2) is: d/dx (1/(x+2)) = -1/(x+2)²
The formula applies to all x except where x = -2, as this would make the denominator zero.
We can derive the derivative of 1/(x+2) using proofs. To show this, we will use algebraic manipulation along with the rules of differentiation.
There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of 1/(x+2) results in -1/(x+2)² using the above-mentioned methods:
The derivative of 1/(x+2) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of 1/(x+2) using the first principle, we will consider f(x) = 1/(x+2). Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)
Given that f(x) = 1/(x+2), we write f(x + h) = 1/(x+h+2).
Substituting these into equation (1), f'(x) = limₕ→₀ [1/(x+h+2) - 1/(x+2)] / h = limₕ→₀ [(x+2) - (x+h+2)] / [(x+2)(x+h+2)h] = limₕ→₀ [-h] / [(x+2)(x+h+2)h] = limₕ→₀ -1 / [(x+2)(x+h+2)]
As h approaches 0, we simplify the expression: f'(x) = -1 / (x+2)²Hence, proved.
To prove the differentiation of 1/(x+2) using the chain rule, We use the formula: 1/(x+2) = (x+2)⁻¹
Let u = x+2
Thus, y = u⁻¹
Using the chain rule formula: d/dx [uⁿ] = nuⁿ⁻¹ du/dx dy/dx = -1 * (x+2)⁻² * d/dx (x+2) dy/dx = -1/(x+2)² Hence, proved.
We will now prove the derivative of 1/(x+2) using the power rule. The step-by-step process is demonstrated below:
Rewrite the function: y = (x+2)⁻¹
Using the power rule to differentiate: d/dx (y) = -1 * (x+2)⁻² * d/dx (x+2) d/dx (y) = -1/(x+2)²
Thus, the derivative of 1/(x+2) is -1/(x+2)².
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky.
To understand them better, think of acceleration (second derivative) as the rate of change of speed (first derivative). Higher-order derivatives make it easier to understand functions like 1/(x+2).
For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.
For the nth Derivative of 1/(x+2), we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change. (continuing for higher-order derivatives).
When x is -2, the derivative is undefined because 1/(x+2) has a vertical asymptote there. When x is 0, the derivative of 1/(x+2) = -1/(0+2)², which is -1/4.
Students frequently make mistakes when differentiating 1/(x+2). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of (1/(x+2) + x²)
Here, we have f(x) = 1/(x+2) + x².
Using the sum rule, differentiate each term separately: The derivative of 1/(x+2) is -1/(x+2)². The derivative of x² is 2x.
Combining these, f'(x) = -1/(x+2)² + 2x.
Thus, the derivative of the specified function is -1/(x+2)² + 2x.
We find the derivative of the given function by separating it into two parts. The first step is finding each derivative and then combining them to get the final result.
A company is analyzing its cost function represented by the function C(x) = 1/(x+2) where C represents the cost at production level x. If x = 3 units, calculate the rate of change of cost.
We have C(x) = 1/(x+2) (cost function)...(1)
Now, we will differentiate the equation (1) Take the derivative of 1/(x+2): dC/dx = -1/(x+2)²
Given x = 3, substitute this into the derivative: dC/dx = -1/(3+2)² = -1/25
Hence, the rate of change of cost at x = 3 units is -1/25.
We calculate the rate of change of cost at x = 3, indicating how cost decreases as production level increases.
Derive the second derivative of the function f(x) = 1/(x+2).
The first step is to find the first derivative, f'(x) = -1/(x+2)²...(1)
Now we will differentiate equation (1) to get the second derivative: f''(x) = d/dx [-1/(x+2)²]
Here we use the chain rule: f''(x) = 2/(x+2)³
Therefore, the second derivative of the function f(x) = 1/(x+2) is 2/(x+2)³.
We use a step-by-step process, where we start with the first derivative. Using the chain rule, we differentiate again to find the second derivative.
Prove: d/dx (1/((x+2)²)) = -2/(x+2)³.
Let’s start using the chain rule: Consider y = 1/((x+2)²) Rewrite as y = (x+2)⁻²
To differentiate, we use the chain rule: dy/dx = -2 * (x+2)⁻³ * d/dx (x+2)
Since d/dx (x+2) = 1, dy/dx = -2/(x+2)³
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation, replacing the function with its derivative step by step.
Solve: d/dx ((x+2)/(x+3))
To differentiate the function, we use the quotient rule: d/dx ((x+2)/(x+3)) = (d/dx (x+2) * (x+3) - (x+2) * d/dx (x+3))/ (x+3)²
We will substitute d/dx (x+2) = 1 and d/dx (x+3) = 1: = (1 * (x+3) - (x+2) * 1)/ (x+3)² = ((x+3) - (x+2)) / (x+3)² = 1/(x+3)²
Therefore, d/dx ((x+2)/(x+3)) = 1/(x+3)²
In this process, we differentiate the given function using the quotient rule. As a final step, we simplify the equation to obtain the final result.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.