BrightChamps Logo
Login
Creative Math Ideas Image
Live Math Learners Count Icon103 Learners

Last updated on July 22nd, 2025

Math Whiteboard Illustration

Derivative of x^sinx

Professor Greenline Explaining Math Concepts

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.

Derivative of x^sinx for Vietnamese Students
Professor Greenline from BrightChamps

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.

Professor Greenline from BrightChamps

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.

Professor Greenline from BrightChamps

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.

Professor Greenline from BrightChamps

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.

Professor Greenline from BrightChamps

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.

Max Pointing Out Common Math Mistakes

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:

Mistake 1

Red Cross Icon Indicating Mistakes to Avoid in This Math Topic

Ignoring Logarithmic Differentiation

Green Checkmark Icon Indicating Correct Solutions in This Math Topic

Students might overlook using logarithmic differentiation, which is crucial for functions like x^sinx. Ensure you start by taking the natural log of both sides to simplify differentiation.

Mistake 2

Red Cross Icon Indicating Mistakes to Avoid in This Math Topic

Misapplying Product and Chain Rules

Green Checkmark Icon Indicating Correct Solutions in This Math Topic

While differentiating ln(y) = sinx * ln(x), students often misapply the product or chain rules. Carefully apply these rules when differentiating composite or product functions.

Mistake 3

Red Cross Icon Indicating Mistakes to Avoid in This Math Topic

Overlooking Domain Restrictions

Green Checkmark Icon Indicating Correct Solutions in This Math Topic

For x^sinx, x must be positive. Students might forget this domain restriction, leading to incorrect results. Always consider the domain when differentiating such functions.

Mistake 4

Red Cross Icon Indicating Mistakes to Avoid in This Math Topic

Incorrect Substitution Back

Green Checkmark Icon Indicating Correct Solutions in This Math Topic

After finding dy/dx in terms of y, students sometimes incorrectly substitute y back into the equation. Ensure y is correctly replaced with x^sinx to avoid errors.

Mistake 5

Red Cross Icon Indicating Mistakes to Avoid in This Math Topic

Neglecting Special Cases

Green Checkmark Icon Indicating Correct Solutions in This Math Topic

Forgetfulness regarding special cases like x = 1 or sinx = 0 may lead to oversight in simplifying results. Recognize these cases early to make appropriate simplifications.

arrow-right
Max from BrightChamps Saying "Hey"
Hey!

Examples Using the Derivative of x^sinx

Ray, the Character from BrightChamps Explaining Math Concepts
Max, the Girl Character from BrightChamps

Problem 1

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

Ray, the Boy Character from BrightChamps Saying "Let’s Begin"
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.

Max from BrightChamps Praising Clear Math Explanations
Well explained 👍
Max, the Girl Character from BrightChamps

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.

Ray, the Boy Character from BrightChamps Saying "Let’s Begin"
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.

Max from BrightChamps Praising Clear Math Explanations
Well explained 👍
Max, the Girl Character from BrightChamps

Problem 3

Derive the second derivative of y = x^sinx.

Ray, the Boy Character from BrightChamps Saying "Let’s Begin"
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.

Max from BrightChamps Praising Clear Math Explanations
Well explained 👍
Max, the Girl Character from BrightChamps

Problem 4

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

Ray, the Boy Character from BrightChamps Saying "Let’s Begin"
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.

Max from BrightChamps Praising Clear Math Explanations
Well explained 👍
Max, the Girl Character from BrightChamps

Problem 5

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

Ray, the Boy Character from BrightChamps Saying "Let’s Begin"
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).

Max from BrightChamps Praising Clear Math Explanations
Well explained 👍
Ray Thinking Deeply About Math Problems

FAQs on the Derivative of x^sinx

1.Find the derivative of x^sinx.

Math FAQ Answers Dropdown Arrow

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

Math FAQ Answers Dropdown Arrow

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

Math FAQ Answers Dropdown Arrow

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

Math FAQ Answers Dropdown Arrow

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

Math FAQ Answers Dropdown Arrow

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

Math FAQ Answers Dropdown Arrow
Professor Greenline from BrightChamps

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.

Math Teacher Background Image
Math Teacher Image

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.

Max, the Girl Character from BrightChamps

Fun Fact

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

INDONESIA - Axa Tower 45th floor, JL prof. Dr Satrio Kav. 18, Kel. Karet Kuningan, Kec. Setiabudi, Kota Adm. Jakarta Selatan, Prov. DKI Jakarta
INDIA - H.No. 8-2-699/1, SyNo. 346, Rd No. 12, Banjara Hills, Hyderabad, Telangana - 500034
SINGAPORE - 60 Paya Lebar Road #05-16, Paya Lebar Square, Singapore (409051)
USA - 251, Little Falls Drive, Wilmington, Delaware 19808
VIETNAM (Office 1) - Hung Vuong Building, 670 Ba Thang Hai, ward 14, district 10, Ho Chi Minh City
VIETNAM (Office 2) - 143 Nguyễn Thị Thập, Khu đô thị Him Lam, Quận 7, Thành phố Hồ Chí Minh 700000, Vietnam
UAE - BrightChamps, 8W building 5th Floor, DAFZ, Dubai, United Arab Emirates
UK - Ground floor, Redwood House, Brotherswood Court, Almondsbury Business Park, Bristol, BS32 4QW, United Kingdom