Summarize this article:
Last updated on September 10, 2025
We use the derivative of e^9x, which is 9e^9x, as a measuring tool for how the exponential 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 e^9x in detail.
We now understand the derivative of e^9x. It is commonly represented as d/dx (e^9x) or (e^9x)', and its value is 9e^9x. The function e^9x has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Exponential Function: (e^9x) is an exponential function where the base is the mathematical constant e and the exponent is 9x.
Chain Rule: Rule for differentiating e^9x since the exponent involves a linear function of x.
Constant Multiple: The derivative involves multiplying by the constant 9 due to the linear function 9x.
The derivative of e^9x can be denoted as d/dx (e^9x) or (e^9x)'.
The formula we use to differentiate e^9x is: d/dx (e^9x) = 9e^9x (or) (e^9x)' = 9e^9x
The formula applies to all x, as exponential functions are differentiable everywhere.
We can derive the derivative of e^9x using proofs. To show this, we will use the chain rule 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 e^9x results in 9e^9x using the above-mentioned methods:
To prove the differentiation of e^9x using the chain rule, We use the formula: Let y = e^9x To differentiate, we consider y = e^u where u = 9x. By the chain rule: dy/dx = dy/du * du/dx We know, dy/du = e^u = e^9x and du/dx = 9
Therefore, dy/dx = e^9x * 9 = 9e^9x
Hence, proved.
Using the properties of exponential functions
The derivative of e^x with respect to x is e^x, which is a fundamental property of exponential functions.
For e^9x, we apply the chain rule as follows: Let y = e^9x
Using the chain rule: dy/dx = 9e^9x
Here, we multiply by the derivative of 9x, which is 9, leading to the result 9e^9x.
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 e^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 e^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).
When x approaches infinity, the function e^9x grows exponentially. When x is zero, the derivative of e^9x = 9e^0, which is 9.
Students frequently make mistakes when differentiating e^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 (e^9x · x^2)
Here, we have f(x) = e^9x · x².
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^9x and v = x².
Let’s differentiate each term, u′= d/dx (e^9x) = 9e^9x v′= d/dx (x²) = 2x
substituting into the given equation, f'(x) = (9e^9x)(x²) + (e^9x)(2x)
Let’s simplify terms to get the final answer, f'(x) = 9x²e^9x + 2xe^9x
Thus, the derivative of the specified function is 9x²e^9x + 2xe^9x.
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.
The temperature of a chemical reaction is modeled by T = e^9x, where T is the temperature in degrees Celsius, and x is time in hours. Calculate the rate of temperature change at x = 1 hour.
We have T = e^9x (rate of temperature change)...(1)
Now, we will differentiate the equation (1) Take the derivative e^9x: dT/dx = 9e^9x
Given x = 1 (substitute this into the derivative) dT/dx = 9e^(9*1) dT/dx = 9e^9
Hence, the rate of temperature change at x = 1 hour is 9e^9 degrees Celsius per hour.
We find the rate of temperature change at x = 1 hour as 9e^9, which means that at this point, the temperature is increasing rapidly.
Derive the second derivative of the function y = e^9x.
The first step is to find the first derivative, dy/dx = 9e^9x...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [9e^9x]
Here we use the constant multiple rule, d²y/dx² = 9 * d/dx [e^9x] d²y/dx² = 9 * 9e^9x d²y/dx² = 81e^9x
Therefore, the second derivative of the function y = e^9x is 81e^9x.
We use the step-by-step process, where we start with the first derivative. Using the constant multiple rule, we differentiate again. We then multiply the constant and simplify the terms to find the final answer.
Prove: d/dx (e^(9x^2)) = 18xe^(9x^2).
Let’s start using the chain rule: Consider y = e^(9x²)
To differentiate, we use the chain rule: dy/dx = d/du [e^u] * du/dx, where u = 9x² dy/dx = e^(9x²) * d/dx [9x²] dy/dx = e^(9x²) * 18x
Substituting y = e^(9x²), d/dx (e^(9x²)) = 18xe^(9x²) Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace the inner function with its derivative. As a final step, we substitute back to derive the equation.
Solve: d/dx (e^9x/x)
To differentiate the function, we use the quotient rule: d/dx (e^9x/x) = (d/dx (e^9x) * x - e^9x * d/dx(x)) / x²
We will substitute d/dx (e^9x) = 9e^9x and d/dx (x) = 1 (9e^9x * x - e^9x * 1) / x² = (9xe^9x - e^9x) / x² = e^9x(9x - 1) / x²
Therefore, d/dx (e^9x/x) = e^9x(9x - 1) / x²
In this process, we differentiate the given function using the product rule and 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.