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114 LearnersLast updated on September 10, 2025
Derivative of 2^-x
We use the derivative of 2^-x to understand how the function changes with respect to x, which can be applied in various real-world contexts. Derivatives are essential for calculating rates of change and optimizing functions. We will now discuss the derivative of 2^-x in detail.
What is the Derivative of 2^-x?
The derivative of 2^-x is commonly represented as d/dx (2^-x) or (2^-x)'. The value of the derivative is -ln(2) * 2^-x. The function 2^-x is differentiable across its domain, and understanding its derivative involves recognizing the exponential nature of the function. The key concepts are mentioned below:
Exponential Function: 2^-x is a type of exponential function.
Natural Logarithm: ln(2) is used in the differentiation of exponential functions with base 2.
Chain Rule: A differentiation rule used to handle composite functions like 2^-x.
Derivative of 2^-x Formula
The derivative of 2^-x can be denoted as d/dx (2^-x) or (2^-x)'.
The formula used to differentiate 2^-x is: d/dx (2^-x) = -ln(2) * 2^-x This formula applies to all x values in the domain of the function.
Proofs of the Derivative of 2^-x
The derivative of 2^-x can be derived using several methods, such as: Using the Chain Rule Using the Definition of Derivative Using the Chain Rule To prove the differentiation of 2^-x using the chain rule, we start by expressing the function as: f(x) = 2^-x = e^(-x ln(2)) Differentiating using the chain rule, we have: f'(x) = d/dx [e^(-x ln(2))] By the chain rule: f'(x) = e^(-x ln(2)) * d/dx (-x ln(2)) = -ln(2) * e^(-x ln(2)) = -ln(2) * 2^-x Hence, proved. Using the Definition of Derivative We can also use the definition of the derivative to prove it: f(x) = 2^-x f'(x) = limₕ→₀ [f(x + h) - f(x)] / h = limₕ→₀ [2^-(x + h) - 2^-x] / h = limₕ→₀ [2^-x (2^-h - 1)] / h Using the identity 2^-h ≈ 1 - h ln(2) when h is small, we have: f'(x) = 2^-x * (-ln(2)) = -ln(2) * 2^-x Thus, the derivative is -ln(2) * 2^-x.
Higher-Order Derivatives of 2^-x
When a function is differentiated multiple times, the results are known as higher-order derivatives. These derivatives can provide insights into the behavior of the function, such as its concavity and rate of change of the rate of change.
For the nth derivative of 2^-x, we use a similar process repeatedly. The first derivative of 2^-x is -ln(2) * 2^-x, indicating how the function decreases. The second derivative is obtained by differentiating the first derivative, leading to (ln(2))^2 * 2^-x.
The pattern continues for higher-order derivatives, with each derivative involving higher powers of ln(2).
Special Cases:
When x=0, the derivative of 2^-x is -ln(2), since 2^0 = 1.
As x approaches infinity, the derivative approaches zero because 2^-x tends to zero.
Common Mistakes and How to Avoid Them in Derivatives of 2^-x
Students often make mistakes when differentiating 2^-x. These errors can be resolved by understanding the correct approach. Here are a few common mistakes and solutions:
Mistake 1
Ignoring the Negative Sign
Students might forget the negative sign in the derivative, leading to an incorrect result. The negative sign arises from the chain rule and should not be omitted. Always ensure that -ln(2) is included in the derivative.
Mistake 2
Misapplying the Chain Rule
Some students might not apply the chain rule correctly, especially when dealing with composite functions. Remember that the derivative of e^(u) is e^(u)u', and apply the chain rule properly to differentiate 2^-x.
Mistake 3
Confusion with Logarithms
Confusion can arise when dealing with logarithms, particularly with ln(2). Ensure you understand the role of ln(2) in the differentiation process and do not confuse it with other logarithmic properties.
Mistake 4
Forgetting the Exponential Base
Students sometimes forget that 2^-x can be rewritten as e^(-x ln(2)). This step is crucial for using the chain rule effectively. Always convert to the exponential form for easier differentiation.
Mistake 5
Incorrect Simplification
In some cases, students may simplify incorrectly, leading to wrong answers. Be careful with algebraic manipulations, especially when simplifying expressions involving exponents and logarithms.
Examples Using the Derivative of 2^-x
Problem 1
Calculate the derivative of 2^-x * ln(x).
Let f(x) = 2^-x * ln(x). Using the product rule, where u = 2^-x and v = ln(x), we get: f'(x) = u'v + uv' u' = -ln(2) * 2^-x v' = 1/x
Substituting these into the product rule: f'(x) = (-ln(2) * 2^-x) * ln(x) + 2^-x * (1/x) = -ln(2) * 2^-x * ln(x) + 2^-x/x
Therefore, the derivative is -ln(2) * 2^-x * ln(x) + 2^-x/x.
Explanation
The derivative is found by applying the product rule to the function, considering both terms and their respective derivatives. We then combine these to arrive at the final result.
Problem 2
A population of bacteria decreases over time according to the function P(t) = 2^-t. If t = 5 hours, find the rate of change of the population.
Given P(t) = 2^-t, we need to find P'(t) at t = 5. P'(t) = d/dt (2^-t) = -ln(2) * 2^-t
Substitute t = 5: P'(5) = -ln(2) * 2^-5 = -ln(2) / 32
Thus, the rate of change of the population at t = 5 is -ln(2) / 32.
Explanation
The rate of change is found by differentiating the population function and substituting the given time to find the specific rate at that moment.
Problem 3
Derive the second derivative of the function y = 2^-x.
The first derivative is: dy/dx = -ln(2) * 2^-x
Now find the second derivative:
d^2y/dx^2 = d/dx [-ln(2) * 2^-x] = -ln(2) * d/dx (2^-x) = -ln(2) * (-ln(2) * 2^-x) = (ln(2))^2 * 2^-x
Therefore, the second derivative is (ln(2))^2 * 2^-x.
Explanation
We start from the first derivative and differentiate again, applying the same principles to find the second derivative, which involves squaring ln(2).
Problem 4
Prove: d/dx (2^-2x) = -2ln(2) * 2^-2x.
Consider y = 2^-2x. Rewrite y as: y = (2^-x)^2
Using the chain rule and power rule:
dy/dx = 2 * (2^-x) * d/dx (2^-x) = 2 * 2^-x * (-ln(2) * 2^-x) = -2ln(2) * 2^-2x
Hence, proved.
Explanation
In this step-by-step process, we use the chain rule to differentiate the equation. We apply the power rule to the outer function and the chain rule to the inner function, leading to the final result.
Problem 5
Solve: d/dx (x * 2^-x).
To differentiate the function, use the product rule:
Let u = x and v = 2^-x. u' = 1 and v' = -ln(2) * 2^-x. Using the product rule:
d/dx (x * 2^-x) = u'v + uv' = (1 * 2^-x) + (x * -ln(2) * 2^-x) = 2^-x - x ln(2) * 2^-x
Therefore, d/dx (x * 2^-x) = 2^-x (1 - x ln(2)).
Explanation
We differentiate the function using the product rule, combining the derivatives of both components, and simplifying the expression to get the final derivative.
FAQs on the Derivative of 2^-x
1.Find the derivative of 2^-x.
2.Can we use the derivative of 2^-x in real life?
3.Is it possible to take the derivative of 2^-x at x = 0?
4.What rule is used to differentiate x * 2^-x?
5.Are the derivatives of 2^-x and 2^x the same?
6.Can we find the derivative of the 2^-x formula?
Important Glossaries for the Derivative of 2^-x
- Derivative: A measure of how a function changes as its input changes, indicating the function's rate of change.
- Exponential Function: A function of the form a^x, where a is a constant and x is the variable.
- Chain Rule: A fundamental rule in calculus used to differentiate composite functions.
- Logarithm: The inverse operation to exponentiation, commonly used in differentiation of exponential functions.
- Product Rule: A rule used to find the derivative of the product of two functions.
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Jaskaran Singh Saluja
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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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