Derivative of x^sinx
2026-02-28 06:05 Diff
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Last updated on August 5, 2025

We explore the derivative of x^sinx, a function that combines both exponential and trigonometric components. Understanding derivatives like this helps us calculate rates of change and solve complex problems in real-life scenarios. We will now delve into the derivative of x^sinx in detail.

What is the Derivative of x^sinx?

The derivative of x^sinx is determined through a combination of differentiation rules, particularly using logarithmic differentiation due to its complexity. The derivative is not straightforward due to the power of x being a function of x. Here are the key concepts involved: - Exponential Function: Functions where a constant is raised to a variable power. - Logarithmic Differentiation: A technique used for differentiating functions of the form u^v, where both u and v are functions of x. - Chain Rule: Used for differentiating composite functions.

Derivative of x^sinx Formula

The derivative of x^sinx can be found by first taking the natural logarithm of both sides and then differentiating: y = x^sinx Take ln of both sides: ln(y) = sinx * ln(x) Differentiate using the product and chain rules: d/dx [ln(y)] = d/dx [sinx * ln(x)] (1/y) * dy/dx = cosx * ln(x) + (sinx/x) Thus, dy/dx = y * [cosx * ln(x) + (sinx/x)] Substitute y = x^sinx back in: dy/dx = x^sinx [cosx * ln(x) + (sinx/x)] This formula is valid for x > 0.

Proofs of the Derivative of x^sinx

We can derive the derivative of x^sinx using logarithmic differentiation. Here's a step-by-step proof: Using Logarithmic Differentiation 1. Start with y = x^sinx. 2. Take the natural log of both sides: ln(y) = sinx * ln(x). 3. Differentiate both sides with respect to x: d/dx [ln(y)] = d/dx [sinx * ln(x)] 4. Apply the chain rule on the left and product rule on the right: (1/y) * dy/dx = cosx * ln(x) + (sinx/x) 5. Multiply both sides by y to get dy/dx: dy/dx = y * [cosx * ln(x) + (sinx/x)] 6. Substitute y = x^sinx back: dy/dx = x^sinx [cosx * ln(x) + (sinx/x)] This proves the derivative of x^sinx.

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Higher-Order Derivatives of x^sinx

Finding higher-order derivatives of x^sinx involves repeated differentiation and can become increasingly complex. Each subsequent derivative requires careful application of the product, chain, and power rules. For the first derivative of a function, we denote it as f′(x), showing how the function changes at a given point. The second derivative, denoted as f′′(x), is derived from the first derivative, indicating the rate of change of the rate of change. Higher-order derivatives like the third, f′′′(x), and beyond continue this pattern and provide deeper insights into the function's behavior.

Special Cases:

When x is 1, the derivative simplifies significantly because x^sinx becomes 1^sinx = 1, so the derivative is 0. When sinx is 0 (x = nπ, where n is an integer), the term cosx * ln(x) will determine the derivative's behavior because sinx/x will be zero.

Common Mistakes and How to Avoid Them in Derivatives of x^sinx

Differentiating x^sinx can be tricky, and students often encounter errors. Understanding the correct approach can help avoid these mistakes. Below are some common errors and tips to resolve them:

Problem 1

Calculate the derivative of x^sinx at x = e.

Okay, lets begin

We have y = x^sinx. Using the derived formula: dy/dx = x^sinx [cosx * ln(x) + (sinx/x)] Substitute x = e: dy/dx = e^sin(e) [cos(e) * ln(e) + (sin(e)/e)] Since ln(e) = 1, dy/dx = e^sin(e) [cos(e) + (sin(e)/e)]. This is the derivative of x^sinx at x = e.

Explanation

The derivative is calculated by substituting x = e into the derived formula. The natural logarithm simplifies since ln(e) = 1, making the process straightforward.

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Problem 2

A company models its production growth over time with P(t) = t^sin(t) for t in years. Find the rate of change of production at t = 1 year.

Okay, lets begin

Given P(t) = t^sin(t), dP/dt = t^sin(t) [cos(t) * ln(t) + (sin(t)/t)] Substitute t = 1: dP/dt = 1^sin(1) [cos(1) * ln(1) + sin(1)] Since ln(1) = 0, dP/dt = sin(1) Thus, the rate of change of production at t = 1 year is sin(1).

Explanation

The derivative simplifies significantly at t = 1 since ln(1) = 0. The expression reduces to sin(1), indicating the rate of change of production at that time.

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Problem 3

Derive the second derivative of y = x^sinx.

Okay, lets begin

First, find the first derivative: dy/dx = x^sinx [cosx * ln(x) + (sinx/x)] Let u = x^sinx and v = cosx*ln(x) + (sinx/x). Find the second derivative using the product rule: d²y/dx² = d/dx[u*v] = u'v + uv' = [x^sinx {cosx * ln(x) + (sinx/x)}]' + x^sinx * [d/dx {cosx * ln(x) + (sinx/x)}] The second derivative involves further differentiation of each term.

Explanation

Finding the second derivative requires applying the product rule to the expression derived for the first derivative. This involves differentiating each component separately.

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Problem 4

Prove: d/dx [(x^sinx)^2] = 2x^sinx * [cosx * ln(x) + (sinx/x)] * x^sinx.

Okay, lets begin

Let y = (x^sinx)^2. Using the chain rule, dy/dx = 2(x^sinx) * d/dx[x^sinx] = 2(x^sinx) * [x^sinx * (cosx * ln(x) + (sinx/x))] = 2x^sinx * [cosx * ln(x) + (sinx/x)] * x^sinx. Hence proved.

Explanation

The chain rule is applied, first differentiating the outer function (square) and then multiplying by the derivative of the inner function (x^sinx), demonstrating the relationship.

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Problem 5

Solve: d/dx (sin(x^sinx)).

Okay, lets begin

To differentiate sin(x^sinx), use the chain rule: d/dx [sin(x^sinx)] = cos(x^sinx) * d/dx [x^sinx] = cos(x^sinx) * [x^sinx * (cosx * ln(x) + (sinx/x))] Thus, d/dx [sin(x^sinx)] = cos(x^sinx) * x^sinx * [cosx * ln(x) + (sinx/x)].

Explanation

The chain rule helps differentiate the composite function, starting with the outer function (sin) and multiplying by the derivative of the inner function (x^sinx).

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FAQs on the Derivative of x^sinx

1.Find the derivative of x^sinx.

Using logarithmic differentiation, the derivative of x^sinx is x^sinx [cosx * ln(x) + (sinx/x)].

2.How can the derivative of x^sinx be applied in real life?

Derivatives like that of x^sinx can be used to model growth rates and changes in systems that involve both exponential and trigonometric relationships, such as in economics and physics.

3.Is the derivative of x^sinx defined for all x?

No, the derivative is only defined for x > 0 because of the logarithmic function involved in the differentiation process.

4.What rule is primarily used to differentiate x^sinx?

Logarithmic differentiation is primarily used, combined with the product and chain rules, to differentiate x^sinx.

5.How does the derivative of x^sinx differ from the derivative of sin(x^x)?

The derivative of x^sinx involves differentiating an exponential function with a trigonometric power, while sin(x^x) involves differentiating a sine function with an exponential argument. The methods and results differ significantly.

6.Can you find the derivative of x^sinx without logarithmic differentiation?

Logarithmic differentiation simplifies the process, but theoretically, applying the product and chain rules directly could work, albeit more complexly. Logarithmic differentiation is the preferred method.

Important Glossaries for the Derivative of x^sinx

Derivative: Indicates how a function changes as its input changes. Logarithmic Differentiation: A technique for differentiating functions with variable exponents. Chain Rule: A rule for differentiating composite functions. Exponential Function: A function where a constant is raised to a variable power. Product Rule: A rule used to differentiate the product of two functions.

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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.

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: He loves to play the quiz with kids through algebra to make kids love it.