Last updated on July 25th, 2025
We use the derivative of 9x, which is 9, as a measuring tool for how the linear function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of 9x in detail.
We now understand the derivative of 9x. It is commonly represented as d/dx (9x) or (9x)', and its value is 9.
The function 9x has a clearly defined derivative, indicating it is differentiable across all real numbers.
The key concepts are mentioned below: Linear Function: A function of the form f(x) = ax + b.
Constant Rule: The derivative of a constant is zero.
Coefficient Rule: The derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function.
The derivative of 9x can be denoted as d/dx (9x) or (9x)'.
The formula we use to differentiate 9x is: d/dx (9x) = 9 The formula applies to all values of x.
We can derive the derivative of 9x using basic calculus rules. To show this, we will use the fundamental rules of differentiation.
There are several methods we use to prove this, such as:
By Definition Using Constant Rule Using Coefficient Rule We will now demonstrate that the differentiation of 9x results in 9 using the above-mentioned methods:
By Definition The derivative of 9x can be proved using the definition of the derivative, which expresses the derivative as the limit of the difference quotient. To find the derivative of 9x using the definition, we will consider f(x) = 9x.
Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 9x, we write f(x + h) = 9(x + h).
Substituting these into equation (1), f'(x) = limₕ→₀ [9(x + h) - 9x] / h = limₕ→₀ [9x + 9h - 9x] / h = limₕ→₀ [9h] / h = limₕ→₀ 9 = 9 Hence, proved. Using Constant Rule To prove the differentiation of 9x using the constant rule, We use the formula: d/dx (c*f(x)) = c * f'(x) Here, c = 9 and f(x) = x Since the derivative of f(x) = x is 1, d/dx (9x) = 9 * 1 = 9
Using Coefficient Rule We will now prove the derivative of 9x using the coefficient rule.
The step-by-step process is demonstrated below: Given that, u(x) = x and c = 9
Using the coefficient rule formula: d/dx [c*u(x)] = c * u'(x) u'(x) = d/dx (x) = 1 Thus, d/dx (9x) = 9 * 1 = 9.
Hence, proved.
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 a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes.
Higher-order derivatives make it easier to understand functions like 9x. 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 9x, 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).
The function 9x is a linear function and does not have special cases of undefined points or asymptotes. For any value of x, the derivative of 9x remains 9.
Students frequently make mistakes when differentiating 9x.
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 (9x · 5x).
Here, we have f(x) = 9x · 5x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 9x and v = 5x. Let’s differentiate each term, u′ = d/dx (9x) = 9 v′ = d/dx (5x) = 5
Substitute these into the given equation, f'(x) = (9) · (5x) + (9x) · (5)
Simplifying terms gives us the final answer, f'(x) = 45x + 45. Thus, the derivative of the specified function is 45x + 45.
We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.
A company calculates the cost of production using the function y = 9x, where y represents the cost for producing x items. If x = 100 items, determine the rate of change of cost.
We have y = 9x (cost function)...(1) Now, we will differentiate the equation (1) Take the derivative: dy/dx = 9
Given x = 100 (substitute this into the derivative)
The rate of change of cost remains constant at 9, regardless of the value of x.
Hence, the rate of change of cost for producing 100 items is 9.
We find the rate of change of cost using the derivative of the cost function, which remains constant at 9 for any number of items produced.
Derive the second derivative of the function y = 9x.
The first step is to find the first derivative, dy/dx = 9...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [9] Since the derivative of a constant is 0, d²y/dx² = 0. Therefore, the second derivative of the function y = 9x is 0.
We use the step-by-step process, where we start with the first derivative. Since the first derivative is a constant, the second derivative is zero.
Prove: d/dx (9x²) = 18x.
Let’s start using the power rule: Consider y = 9x² To differentiate, we use the power rule: dy/dx = 9 * d/dx [x²] Since the derivative of x² is 2x, dy/dx = 9 * 2x Therefore, d/dx (9x²) = 18x. Hence proved.
In this step-by-step process, we used the power rule to differentiate the equation. Then, we replace x² with its derivative. As a final step, we substitute and simplify to derive the equation.
Solve: d/dx (9x/x).
To differentiate the function, we first simplify it: d/dx (9x/x) = d/dx (9) Since 9 is a constant, its derivative is 0. Therefore, d/dx (9x/x) = 0.
In this process, we simplify the given function and use the differentiation rule for constants 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.
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