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Last updated on September 15, 2025
We use the derivative of x^y to understand how this power function changes in response to a slight change in x. Derivatives are crucial in various fields such as economics and physics for calculating predictions and analyzing trends. We will now explore the derivative of x^y in detail.
The derivative of xy can be expressed as d/dx (xy) or (xy)'.
When y is a constant, the derivative involves the power rule.
However, when y is a function of x, the derivative becomes more complex.
The main concepts are:
Power Function: xy, where y can be a constant or a function of x.
Chain Rule: Used when y is a function of x.
Logarithmic Differentiation: Useful for differentiating xy when y is not constant.
The derivative of xy can be determined using logarithmic differentiation when y is a function of x. The formula we use is: d/dx (xy) = yx(y-1) + xy ln(x) dy/dx This formula applies to all x > 0.
We can derive the derivative of xy using several methods. Here are some approaches:
Logarithmic Differentiation
Using the Chain Rule Let's demonstrate the differentiation of xy:
Logarithmic Differentiation Consider f(x) = xy, where y = g(x).
Take the natural logarithm of both sides: ln f(x) = y ln(x)
Differentiate both sides with respect to x: (1/f(x)) f'(x) = y (1/x) + ln(x) dy/dx
Multiply through by f(x) = xy: f'(x) = xy (y/x + ln(x) dy/dx)
Simplifying, we find: f'(x) = y x(y-1) + xy ln(x) dy/dx
Using the Chain Rule For xy with y as a constant, we use the power rule: y x(y-1)
If y is a function of x, we apply the chain rule: f(x) = xg(x)
Let u = g(x) f'(x) = d/dx(xu) = xu ln(x) du/dx
Higher-order derivatives involve differentiating a function multiple times.
For xy, each differentiation can introduce complexity, especially if y is a function of x.
The first derivative, f′(x), gives the rate of change of the function.
The second derivative, f′′(x), provides the rate of change of the first derivative.
For the nth Derivative of xy, we use fⁿ(x) for the nth derivative of a function f(x), indicating the change in the rate of change.
When y is a constant, the derivative simplifies to the power rule. When x = 1, the derivative simplifies since xy becomes 1y, which is 1.
Students often make mistakes when differentiating xy, especially when y is a function of x. These mistakes can be minimized by following proper steps. Here are a few common mistakes and how to avoid them:
Calculate the derivative of (x^y · ln(x))
Here, we have f(x) = xy · ln(x).
Using the product rule, f'(x) = u′v + uv′ In this equation, u = xy and v = ln(x).
Differentiate each term: u′ = y x(y-1) + xy ln(x) dy/dx v′ = 1/x
Substitute into the equation: f'(x) = (y x(y-1) + xy ln(x) dy/dx) ln(x) + xy (1/x)
Simplify to get the final answer: f'(x) = y x(y-1) ln(x) + xy ln(x) dy/dx + x(y-1)
Thus, the derivative is y x(y-1) ln(x) + xy ln(x) dy/dx + x(y-1).
We find the derivative by dividing the function into two parts and applying the product rule. The first step is differentiating each term, then combining them for the final result.
A company uses a model where the revenue R is given by R = x^y dollars, where x is the number of products sold, and y is determined by market conditions. If x = 2 and y = 3, find the rate of change of revenue with respect to x.
We have R = xy... (1)
Differentiate (1) with respect to x:
dR/dx = y x(y-1)
Given x = 2 and y = 3, substitute these values:
dR/dx = 3 (2)(3-1) = 3 * 4 = 12
Hence, the rate of change of revenue with respect to the number of products sold is 12 dollars per product.
In this example, we differentiate the revenue function with respect to the number of products sold. Substituting x and y, we calculate the rate of change in revenue.
Derive the second derivative of the function f(x) = x^y, where y = g(x).
First, find the first derivative:
f'(x) = y x(y-1) + xy ln(x) dy/dx... (1)
Differentiate equation (1) to get the second derivative:
f''(x) = d/dx [y x(y-1) + xy ln(x) dy/dx]
Use the product rule: f''(x) = y (y-1) x(y-2) + d/dx [xy ln(x) dy/dx]
Apply the product rule to the second term: f''(x) = y (y-1) x(y-2) + ln(x) (xy) dy/dx + xy d/dx [ln(x) dy/dx]
Simplify to get the final answer.
We obtain the second derivative by first finding the first derivative and then applying the product rule again to differentiate the result.
Prove: d/dx (x^(2y)) = 2y x^(2y-1) + x^(2y) ln(x) 2 dy/dx.
Let's use logarithmic differentiation:
Consider f(x) = x(2y)
Take the natural logarithm: ln f(x) = 2y ln(x)
Differentiate both sides: 1/f(x) f'(x) = 2y (1/x) + 2 ln(x) dy/dx
Multiply through by f(x) = x(2y): f'(x) = 2y x^(2y-1) + x^(2y) ln(x) 2 dy/dx
Hence proved.
We use logarithmic differentiation to differentiate x(2y). By taking the natural logarithm and differentiating both sides, we derive the expression.
Solve: d/dx (x^y/x)
To differentiate the function, use the quotient rule:
d/dx (xy/x) = (d/dx (xy) * x - x^y * d/dx(x))/x2
Substitute d/dx (xy) = y x(y-1) + xy ln(x)
dy/dx: = (y x(y-1) + xy ln(x) dy/dx) * x - xy * 1 / x2 = (y xy + x(y+1) ln(x) dy/dx - xy) / x2 = (y xy - x^y + x(y+1) ln(x) dy/dx) / x2
Therefore, d/dx (xy/x) = (y xy - xy + x(y+1) ln(x) dy/dx) / x2
In this process, we differentiate the given function using the quotient rule. We then simplify the equation to find 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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