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Last updated on October 8, 2025
We use the derivative of x^(1/x) to understand how this 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 x^(1/x) in detail.
We now understand the derivative of x(1/x). It is commonly represented as d/dx (x(1/x)) or (x(1/x))', and its value is a bit more complex. The function x(1/x) has a defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Power Function: x(1/x) can be rewritten as e(ln(x)/x).
Chain Rule: Rule for differentiating functions like e(ln(x)/x).
Logarithmic Differentiation: Useful for differentiating functions like x(1/x).
The derivative of x(1/x) can be denoted as d/dx (x(1/x)) or (x(1/x))'.
The formula we use to differentiate x(1/x) involves logarithmic differentiation: d/dx (x(1/x)) = x(1/x) * [(1 - ln(x))x²]
This formula applies to all x > 0.
We can derive the derivative of x(1/x) using proofs. To show this, we will use logarithmic differentiation 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 x(1/x) results in x(1/x) * [(1 - ln(x))x²] using the above-mentioned methods:
The derivative of x(1/x) can be proved using logarithmic differentiation, which simplifies the differentiation process for functions raised to variable exponents. To find the derivative of x(1/x), consider y = x(1/x). Taking the natural logarithm on both sides, ln(y) = (1/x) * ln(x) Differentiating both sides with respect to x, 1/y * dy/dx = (d/dx (ln(x))x) = (1/x²) - (ln(x)x²) dy/dx = y * [(1 - ln(x))x²] Since y = x(1/x), substitute back to get the derivative: dy/dx = x(1/x) * [(1 - ln(x))x²]
To prove the differentiation of x(1/x) using the chain rule, We use the transformation: x(1/x) = e(ln(x)/x) Consider u(x) = ln(x)x, then y = eu By chain rule: dy/dx = eu * du/dx du/dx = (1/x²) - (ln(x)/x²) Thus, dy/dx = e(ln(x)/x) * [(1 - ln(x))/x²] Since e(ln(x)/x) = x(1/x), dy/dx = x(1/x) * [(1 - ln(x))/x²]
We can rewrite the function as e^(ln(x)/x) for differentiation. Let u = ln(x) and v = x Therefore, y = u/v Using the quotient rule: d/dx (u/v) = (v * du/dx - u * dv/dx)/v² du/dx = 1/x and dv/dx = 1 Applying to the quotient rule: (1 * 1/x - ln(x) * 1)/x² = (1 - ln(x))/x² Thus, the derivative is: dy/dx = x(1/x) * [(1 - ln(x))/x²]
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 x(1/x).
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 x(1/x), 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 (continuing for higher-order derivatives).
When x is 0, the function x(1/x) is not defined, and thus its derivative is also undefined.
For x = 1, the derivative of x(1/x) simplifies to 1, as x(1/x) = 1 and the derivative of a constant is 0.
Students frequently make mistakes when differentiating x(1/x). 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 (x^(1/x)·x^2)
Here, we have f(x) = x(1/x)·x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x(1/x) and v = x². Let’s differentiate each term, u′ = d/dx (x(1/x)) = x(1/x) * [(1 - ln(x))x²] v′ = d/dx (x²) = 2x Substituting into the given equation, f'(x) = [x(1/x) * [(1 - ln(x))x²]] * x² + x(1/x) * 2x Let’s simplify terms to get the final answer, f'(x) = x(1/x) * (1 - ln(x)) + 2x(1/x+1) Thus, the derivative of the specified function is x(1/x) * (1 - ln(x)) + 2x(1/x+1).
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 balloon is being inflated in the shape of a sphere, and its volume is given by V = x^(1/x) cubic meters, where x is the radius in meters. If the radius is 4 meters, find the rate at which the volume changes with respect to the radius.
We have V = x(1/x) (volume of the balloon)...(1) Now, we will differentiate the equation (1) Take the derivative of x(1/x): dV/dx = x(1/x) * [(1 - ln(x))x²] Given x = 4 (substitute this into the derivative) dV/dx = 4(1/4) * [(1 - ln(4))/16] Calculating further, 4(1/4) = √2, ln(4) ≈ 1.386 dV/dx ≈ √2 * [(1 - 1.386)/16] dV/dx ≈ √2 * [-0.386/16] dV/dx ≈ -0.034 Hence, the rate at which the volume changes with respect to the radius at x = 4 meters is approximately -0.034 cubic meters per meter.
We find the rate of change of volume at x = 4 meters by differentiating the volume function and substituting the given radius to calculate the rate at which the volume changes.
Derive the second derivative of the function y = x^(1/x).
The first step is to find the first derivative, dy/dx = x(1/x) * [(1 - ln(x))x²]...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [x(1/x) * [(1 - ln(x))x²]] Using the product rule, d²y/dx² = [d/dx (x(1/x)) * (1 - ln(x))x²] + [x(1/x) * d/dx ((1 - ln(x))x²)] Calculating each part, d/dx (x(1/x)) = x(1/x) * [(1 - ln(x))x²] d/dx ((1 - ln(x))x²) = [-1/x² - (1/x * -2/x³)] = [-1/x² + 2ln(x)x³] Substitute back, d²y/dx² = [x(1/x) * [(1 - ln(x))x²] * (1 - ln(x))x²] + [x(1/x) * [-1/x² + 2ln(x)x³]] Simplifying further will give the second derivative in more detailed form.
We use the step-by-step process, where we start with the first derivative.
Using the product rule, we differentiate the first derivative to find the second derivative.
We then substitute the identities and simplify the terms to find the final answer.
Prove: d/dx (x^(1/x+1)) = x^(1/x+1) * [(1 - ln(x+1))/(x+1)²].
Let’s start using the chain rule: Consider y = x(1/x+1) Rewrite y = e((ln(x))/(x+1)) Differentiate using the chain rule: dy/dx = e((ln(x))/(x+1)) * d/dx ((ln(x))/(x+1)) The derivative of (ln(x))/(x+1) using quotient rule: d/dx ((ln(x))/(x+1)) = [(x+1)(1/x) - ln(x)(1)]/(x+1)² = [1 - ln(x)]/(x+1)² Substituting back, dy/dx = x(1/x+1) * [(1 - ln(x))/(x+1)²] Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation.
Then, we replace the expression with its derivative.
As a final step, we substitute y = x(1/x+1) to derive the equation.
Solve: d/dx (x^(1/x)/x)
To differentiate the function, we use the quotient rule: d/dx (x(1/x)/x) = (d/dx (x(1/x)) * x - x(1/x) * d/dx(x))x² We will substitute d/dx (x(1/x)) = x(1/x) * [(1 - ln(x))x²] and d/dx (x) = 1 = [x(1/x) * [(1 - ln(x))x²] * x - x(1/x)]/x² = [x(1/x) * (1 - ln(x)) - x(1/x)]/x³ = x(1/x) * [(1 - ln(x)) - 1]/x³ Therefore, d/dx (x(1/x)/x) = x(1/x) * [-ln(x)]/x³
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.
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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