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Last updated on September 26, 2025
We use the derivative of 2^(x^3), which requires the use of chain rule and logarithmic differentiation, as a way to understand how exponential functions change in response to a slight change in x. Derivatives help us calculate growth or decay in real-life situations. We will now talk about the derivative of 2^(x^3) in detail.
We now understand the derivative of 2^(x^3). It is commonly represented as d/dx (2^(x^3)) or (2^(x^3))', and its value is 3x²·2^(x^3)ln(2). The function 2^(x^3) has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Exponential Function: (2^(x^3) is an exponential function with a variable exponent).
Chain Rule: Rule for differentiating functions of functions.
Logarithmic Differentiation: Used for differentiating exponential functions with variable exponents.
The derivative of 2(x3) can be denoted as d/dx (2(x^3)) or (2(x^3))'. The formula we use to differentiate 2(x3) is: d/dx (2(x3)) = 3x²·2(x^3)ln(2)
The formula applies to all x within the domain of the function.
We can derive the derivative of 2(x3) using proofs. To show this, we will use the chain rule and logarithmic differentiation.
There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of 2^(x^3) results in 3x²·2^(x^3)ln(2) using the above-mentioned methods:
The derivative of 2(x3) can be proved using logarithmic differentiation. To find the derivative of 2(x3), we first take the natural log of both sides: Let y = 2(x3). ln(y) = x3·ln(2) Differentiating both sides with respect to x: d/dx [ln(y)] = d/dx [x3·ln(2)] 1/y·dy/dx = 3x²·ln(2) dy/dx = y·3x²·ln(2) Substitute y = 2(x3): dy/dx = 2(x3)·3x²·ln(2) Hence, proved.
To prove the differentiation of 2^(x^3) using the chain rule, consider: Let y = 2^(x^3) Taking the natural log, ln(y) = x^3·ln(2) Differentiating both sides: 1/y·dy/dx = 3x²·ln(2) dy/dx = y·3x²·ln(2) Substitute y = 2^(x^3): dy/dx = 2^(x^3)·3x²·ln(2) Thus, the derivative using the chain rule is 3x²·2^(x^3)ln(2).
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be complex. To understand them better, think of a situation where a quantity grows exponentially, and the rate of growth also changes over time. Higher-order derivatives make it easier to understand functions like 2^(x^3).
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 2^(x^3), we generally use f^(n)(x) for the nth derivative of a function f(x), which tells us the change in the rate of change.
When x = 0, the derivative of 2^(x^3) is 0 because 3x² becomes 0.
When x is very large, the derivative increases rapidly due to the exponential nature of the function.
Students frequently make mistakes when differentiating 2^(x^3). 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 2^(x^3)·x².
Here, we have f(x) = 2^(x^3)·x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 2^(x^3) and v = x². Let’s differentiate each term, u′ = d/dx (2^(x^3)) = 3x²·2^(x^3)ln(2) v′ = d/dx (x²) = 2x Substituting into the given equation, f'(x) = (3x²·2^(x^3)ln(2))·x² + 2^(x^3)·2x Let’s simplify terms to get the final answer, f'(x) = 3x^4·2^(x^3)ln(2) + 2x·2^(x^3) Thus, the derivative of the specified function is 3x^4·2^(x^3)ln(2) + 2x·2^(x^3).
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 tracks its growth using the function y = 2^(x^3), where y represents revenue, and x represents time in years. What is the rate of change of revenue when x = 1 year?
We have y = 2^(x^3) (revenue growth function)...(1) Now, we will differentiate the equation (1) Take the derivative of 2^(x^3): dy/dx = 3x²·2^(x^3)ln(2) Given x = 1, substitute this into the derivative: dy/dx = 3(1)²·2^(1^3)ln(2) = 3·2·ln(2) Hence, we get the rate of change of revenue when x = 1 year as 6ln(2).
We find the rate of change of revenue at x = 1 year by differentiating the function and substituting the given value of x to calculate the result.
Derive the second derivative of the function y = 2^(x^3).
The first step is to find the first derivative, dy/dx = 3x²·2^(x^3)ln(2)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [3x²·2^(x^3)ln(2)] Here we use the product rule, d²y/dx² = 3ln(2)·[2x·2^(x^3) + x²·d/dx(2^(x^3))] = 3ln(2)·[2x·2^(x^3) + x²·3x²·2^(x^3)ln(2)] = 3ln(2)·[2x·2^(x^3) + 3x^4·2^(x^3)ln(2)] Therefore, the second derivative of the function y = 2^(x^3) is 6x·2^(x^3)ln(2) + 9x^4·2^(x^3)ln²(2).
We use the step-by-step process, where we start with the first derivative.
Using the product rule, we differentiate 3x²·2^(x^3)ln(2).
We then substitute the identity and simplify the terms to find the final answer.
Prove: d/dx (2^(2x^3)) = 6x²·2^(2x^3)ln(2).
Let’s start using the chain rule: Consider y = 2^(2x^3) To differentiate, we use the chain rule: ln(y) = 2x^3·ln(2) Differentiating both sides: 1/y·dy/dx = 6x²·ln(2) dy/dx = y·6x²·ln(2) Substituting y = 2^(2x^3), d/dx (2^(2x^3)) = 6x²·2^(2x^3)ln(2) Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation.
Then, we replace 2^(2x^3) with its derivative.
As a final step, we substitute y = 2^(2x^3) to derive the equation.
Solve: d/dx (2^(x^3)/x)
To differentiate the function, we use the quotient rule: d/dx (2^(x^3)/x) = (d/dx (2^(x^3))·x - 2^(x^3)·d/dx(x))/x² We will substitute d/dx (2^(x^3)) = 3x²·2^(x^3)ln(2) and d/dx(x) = 1 = (3x²·2^(x^3)ln(2)·x - 2^(x^3)·1)/x² = (3x³·2^(x^3)ln(2) - 2^(x^3))/x² Therefore, d/dx (2^(x^3)/x) = (3x³·2^(x^3)ln(2) - 2^(x^3))/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.
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