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106 LearnersLast updated on October 3, 2025
Derivative of e^cosx
We use the derivative of e^cosx, which involves the chain rule, as a tool to understand how the function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now discuss the derivative of e^cosx in detail.
What is the Derivative of e^cosx?
We now understand the derivative of e^cosx. It is commonly represented as d/dx (e^cosx) or (e^cosx)', and its value is -sin(x)·ecosx.
The function ecosx has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:
Exponential Function: ex, where e is a constant approximately equal to 2.71828.
Chain Rule: Rule for differentiating composite functions.
Cosine Function: cos(x) is a trigonometric function.
Derivative of e^cosx Formula
The derivative of ecosx can be denoted as d/dx (ecosx) or (ecosx)'.
The formula we use to differentiate e^cosx is: d/dx (ecosx) = -sin(x)·ecosx
The formula applies to all x within the domain of cos(x).
Proofs of the Derivative of e^cosx
We can derive the derivative of e^cosx 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:
- By Using Chain Rule
- Using the properties of exponential functions
- Using Chain Rule
To prove the differentiation of e^cosx using the chain rule, We use the formula: f(x) = e^u(x), where u(x) = cos(x)
The derivative f'(x) is \( e^u(x)·u'(x).
\)
Let’s substitute u(x) = cos(x) and find u'(x) = -sin(x). Thus, f'(x) = e^cosx·(-sin(x)) = -sin(x)·e^cosx
Hence, proved.
Higher-Order Derivatives of e^cosx
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^cosx.
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^cosx, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change.
Special Cases:
When x is π, the derivative is zero because sin(π) = 0. When x is 0, the derivative of e^cosx = -sin(0)·e^1, which is 0.
Common Mistakes and How to Avoid Them in Derivatives of e^cosx
Students frequently make mistakes when differentiating e^cosx. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Mistake 1
Not Applying the Chain Rule
Students often forget to use the chain rule. This happens when the derivative of the inner function is not considered. For example: Incorrect: d/dx (e^cosx) = e^cosx.
To fix this error, students should divide the functions into inner and outer parts.
Then, make sure that each function is differentiated. For example, d/dx (e^cosx) = -sin(x)·e^cosx.
Mistake 2
Ignoring the Negative Sign in the Derivative
Students might not notice the negative sign that results from differentiating cos(x) to -sin(x). Always remember that the derivative of cos(x) is -sin(x).
Mistake 3
Not Simplifying the Equation
Students may forget to simplify the equation, which can lead to incomplete or incorrect results. They often skip steps and directly arrive at the result. Ensure that each step is written in order. Students might think it is awkward, but it is important to avoid errors in the process.
Mistake 4
Confusing Exponential and Trigonometric Derivatives
Students sometimes confuse the rules for differentiating exponential functions with trigonometric ones. For example, assuming that the derivative of e^cosx is similar to that of e^x. Remember that e^x is its own derivative, but when combined with another function like cos(x), the chain rule must be applied.
Mistake 5
Misidentifying the Chain Rule Components
A common mistake is not correctly identifying the inner and outer functions when applying the chain rule. For e^cosx, e^u is the outer function and cosx is the inner function. Ensure both are identified correctly before proceeding with differentiation.
Examples Using the Derivative of e^cosx
Problem 1
Calculate the derivative of (e^cosx·sinx)
Here, we have f(x) = e^cosx·sinx. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = e^cosx and v = sinx.
Let’s differentiate each term, u′ = d/dx (e^cosx) = -sin(x)·e^cosx v′ = d/dx (sinx) = cosx
Substituting into the given equation, f'(x) = (-sin(x)·e^cosx)·(sinx) + (e^cosx)·(cosx)
Let’s simplify terms to get the final answer, f'(x) = -sin^2(x)·e^cosx + cosx·e^cosx
Thus, the derivative of the specified function is e^cosx(cosx - sin^2x).
Explanation
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.
Problem 2
A company observes that the output of a machine follows the function y = e^cosx, where y represents output and x represents time in hours. If x = π/3 hours, determine the rate of change of output.
We have y = e^cosx (output function)...(1)
Now, we will differentiate the equation (1)
Take the derivative of e^cosx: dy/dx = -sin(x)·e^cosx
Given x = π/3 (substitute this into the derivative)
dy/dx = -sin(π/3)·e^cos(π/3) = -√3/2·e^(1/2)
Hence, we get the rate of change of output at x = π/3 hours as -√3/2·e^(1/2).
Explanation
We find the rate of change of output at x = π/3 by substituting the value of x into the derivative. This gives us the rate at which the output is changing at that specific time.
Problem 3
Derive the second derivative of the function y = e^cosx.
The first step is to find the first derivative, dy/dx = -sin(x)·e^cosx...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-sin(x)·e^cosx]
Here we use the product rule, d²y/dx² = [-cos(x)·e^cosx]·(-sin(x)) + [-sin(x)]·[-sin(x)·e^cosx] = cos(x)·sin(x)·e^cosx + sin²(x)·e^cosx
Therefore, the second derivative of the function y = e^cosx is e^cosx[sin²(x) + cos(x)·sin(x)].
Explanation
We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate each component. We then substitute the derived identities and simplify the terms to find the final answer.
Problem 4
Prove: d/dx (e^cos²x) = -2sin(x)cos(x)e^cos²x.
Let’s start using the chain rule: Consider y = e^cos²x = e^(cos(x)²)
To differentiate, we use the chain rule: dy/dx = e^u(x)·u'(x), where u(x) = cos(x)² u'(x) = 2cos(x)·(-sin(x))
Thus, dy/dx = e^cos²x·[-2sin(x)cos(x)] = -2sin(x)cos(x)e^cos²x Hence proved.
Explanation
In this step-by-step process, we used the chain rule to differentiate the equation. We identified the inner function and its derivative, then substituted back to derive the equation.
Problem 5
Solve: d/dx (e^cosx/x)
To differentiate the function, we use the quotient rule: d/dx (e^cosx/x) = (d/dx (e^cosx)·x - e^cosx·d/dx(x))/x²
We will substitute d/dx (e^cosx) = -sin(x)·e^cosx and d/dx (x) = 1 = ([-sin(x)·e^cosx]·x - e^cosx·1)/x² = (-x·sin(x)·e^cosx - e^cosx)/x² = -e^cosx(x·sin(x) + 1)/x²
Therefore, d/dx (e^cosx/x) = -e^cosx(x·sin(x) + 1)/x².
Explanation
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.
FAQs on the Derivative of e^cosx
1.Find the derivative of e^cosx.
2.Can we use the derivative of e^cosx in real life?
3.Is it possible to take the derivative of e^cosx at the point where x = π/2?
4.What rule is used to differentiate e^cosx/x?
5.Are the derivatives of e^cosx and e^-cosx the same?
6.Can we find the derivative of the e^cosx formula?
Important Glossaries for the Derivative of e^cosx
- Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.
- Exponential Function: A function in which a constant base is raised to a variable exponent, commonly written as e^x.
- Chain Rule: A rule used in calculus for differentiating the composition of two or more functions.
- Trigonometric Function: A function related to angles and ratios, such as sine, cosine, and tangent.
- Product Rule: A rule used to differentiate products of two functions, written as d/dx [u·v] = u'v + uv'.
Jaskaran Singh Saluja
About the Author
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.
Fun Fact
: He loves to play the quiz with kids through algebra to make kids love it.
