1 added
2 removed
Original
2026-01-01
Modified
2026-02-28
1
-
<p>218 Learners</p>
1
+
<p>248 Learners</p>
2
<p>Last updated on<strong>August 5, 2025</strong></p>
2
<p>Last updated on<strong>August 5, 2025</strong></p>
3
<p>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.</p>
3
<p>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.</p>
4
<h2>What is the Derivative of x^sinx?</h2>
4
<h2>What is the Derivative of x^sinx?</h2>
5
<p>The derivative of x^sinx is determined through a<a>combination</a>of differentiation rules, particularly using logarithmic differentiation due to its complexity. The derivative is not straightforward due to the<a>power</a>of x being a<a>function</a>of x. Here are the key concepts involved: - Exponential Function: Functions where a<a>constant</a>is raised to a<a>variable</a>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.</p>
5
<p>The derivative of x^sinx is determined through a<a>combination</a>of differentiation rules, particularly using logarithmic differentiation due to its complexity. The derivative is not straightforward due to the<a>power</a>of x being a<a>function</a>of x. Here are the key concepts involved: - Exponential Function: Functions where a<a>constant</a>is raised to a<a>variable</a>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.</p>
6
<h2>Derivative of x^sinx Formula</h2>
6
<h2>Derivative of x^sinx Formula</h2>
7
<p>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<a>product</a>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<a>formula</a>is valid for x > 0.</p>
7
<p>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<a>product</a>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<a>formula</a>is valid for x > 0.</p>
8
<h2>Proofs of the Derivative of x^sinx</h2>
8
<h2>Proofs of the Derivative of x^sinx</h2>
9
<p>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<a>log</a>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.</p>
9
<p>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<a>log</a>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.</p>
10
<h3>Explore Our Programs</h3>
10
<h3>Explore Our Programs</h3>
11
-
<p>No Courses Available</p>
12
<h2>Higher-Order Derivatives of x^sinx</h2>
11
<h2>Higher-Order Derivatives of x^sinx</h2>
13
<p>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<a>rate</a>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.</p>
12
<p>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<a>rate</a>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.</p>
14
<h2>Special Cases:</h2>
13
<h2>Special Cases:</h2>
15
<p>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<a>integer</a>), the<a>term</a>cosx * ln(x) will determine the derivative's behavior because sinx/x will be zero.</p>
14
<p>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<a>integer</a>), the<a>term</a>cosx * ln(x) will determine the derivative's behavior because sinx/x will be zero.</p>
16
<h2>Common Mistakes and How to Avoid Them in Derivatives of x^sinx</h2>
15
<h2>Common Mistakes and How to Avoid Them in Derivatives of x^sinx</h2>
17
<p>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:</p>
16
<p>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:</p>
18
<h3>Problem 1</h3>
17
<h3>Problem 1</h3>
19
<p>Calculate the derivative of x^sinx at x = e.</p>
18
<p>Calculate the derivative of x^sinx at x = e.</p>
20
<p>Okay, lets begin</p>
19
<p>Okay, lets begin</p>
21
<p>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.</p>
20
<p>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.</p>
22
<h3>Explanation</h3>
21
<h3>Explanation</h3>
23
<p>The derivative is calculated by substituting x = e into the derived formula. The natural logarithm simplifies since ln(e) = 1, making the process straightforward.</p>
22
<p>The derivative is calculated by substituting x = e into the derived formula. The natural logarithm simplifies since ln(e) = 1, making the process straightforward.</p>
24
<p>Well explained 👍</p>
23
<p>Well explained 👍</p>
25
<h3>Problem 2</h3>
24
<h3>Problem 2</h3>
26
<p>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.</p>
25
<p>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.</p>
27
<p>Okay, lets begin</p>
26
<p>Okay, lets begin</p>
28
<p>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).</p>
27
<p>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).</p>
29
<h3>Explanation</h3>
28
<h3>Explanation</h3>
30
<p>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.</p>
29
<p>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.</p>
31
<p>Well explained 👍</p>
30
<p>Well explained 👍</p>
32
<h3>Problem 3</h3>
31
<h3>Problem 3</h3>
33
<p>Derive the second derivative of y = x^sinx.</p>
32
<p>Derive the second derivative of y = x^sinx.</p>
34
<p>Okay, lets begin</p>
33
<p>Okay, lets begin</p>
35
<p>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.</p>
34
<p>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.</p>
36
<h3>Explanation</h3>
35
<h3>Explanation</h3>
37
<p>Finding the second derivative requires applying the product rule to the expression derived for the first derivative. This involves differentiating each component separately.</p>
36
<p>Finding the second derivative requires applying the product rule to the expression derived for the first derivative. This involves differentiating each component separately.</p>
38
<p>Well explained 👍</p>
37
<p>Well explained 👍</p>
39
<h3>Problem 4</h3>
38
<h3>Problem 4</h3>
40
<p>Prove: d/dx [(x^sinx)^2] = 2x^sinx * [cosx * ln(x) + (sinx/x)] * x^sinx.</p>
39
<p>Prove: d/dx [(x^sinx)^2] = 2x^sinx * [cosx * ln(x) + (sinx/x)] * x^sinx.</p>
41
<p>Okay, lets begin</p>
40
<p>Okay, lets begin</p>
42
<p>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.</p>
41
<p>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.</p>
43
<h3>Explanation</h3>
42
<h3>Explanation</h3>
44
<p>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.</p>
43
<p>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.</p>
45
<p>Well explained 👍</p>
44
<p>Well explained 👍</p>
46
<h3>Problem 5</h3>
45
<h3>Problem 5</h3>
47
<p>Solve: d/dx (sin(x^sinx)).</p>
46
<p>Solve: d/dx (sin(x^sinx)).</p>
48
<p>Okay, lets begin</p>
47
<p>Okay, lets begin</p>
49
<p>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)].</p>
48
<p>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)].</p>
50
<h3>Explanation</h3>
49
<h3>Explanation</h3>
51
<p>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).</p>
50
<p>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).</p>
52
<p>Well explained 👍</p>
51
<p>Well explained 👍</p>
53
<h2>FAQs on the Derivative of x^sinx</h2>
52
<h2>FAQs on the Derivative of x^sinx</h2>
54
<h3>1.Find the derivative of x^sinx.</h3>
53
<h3>1.Find the derivative of x^sinx.</h3>
55
<p>Using logarithmic differentiation, the derivative of x^sinx is x^sinx [cosx * ln(x) + (sinx/x)].</p>
54
<p>Using logarithmic differentiation, the derivative of x^sinx is x^sinx [cosx * ln(x) + (sinx/x)].</p>
56
<h3>2.How can the derivative of x^sinx be applied in real life?</h3>
55
<h3>2.How can the derivative of x^sinx be applied in real life?</h3>
57
<p>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.</p>
56
<p>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.</p>
58
<h3>3.Is the derivative of x^sinx defined for all x?</h3>
57
<h3>3.Is the derivative of x^sinx defined for all x?</h3>
59
<p>No, the derivative is only defined for x > 0 because of the<a>logarithmic function</a>involved in the differentiation process.</p>
58
<p>No, the derivative is only defined for x > 0 because of the<a>logarithmic function</a>involved in the differentiation process.</p>
60
<h3>4.What rule is primarily used to differentiate x^sinx?</h3>
59
<h3>4.What rule is primarily used to differentiate x^sinx?</h3>
61
<p>Logarithmic differentiation is primarily used, combined with the product and chain rules, to differentiate x^sinx.</p>
60
<p>Logarithmic differentiation is primarily used, combined with the product and chain rules, to differentiate x^sinx.</p>
62
<h3>5.How does the derivative of x^sinx differ from the derivative of sin(x^x)?</h3>
61
<h3>5.How does the derivative of x^sinx differ from the derivative of sin(x^x)?</h3>
63
<p>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<a>argument</a>. The methods and results differ significantly.</p>
62
<p>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<a>argument</a>. The methods and results differ significantly.</p>
64
<h3>6.Can you find the derivative of x^sinx without logarithmic differentiation?</h3>
63
<h3>6.Can you find the derivative of x^sinx without logarithmic differentiation?</h3>
65
<p>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.</p>
64
<p>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.</p>
66
<h2>Important Glossaries for the Derivative of x^sinx</h2>
65
<h2>Important Glossaries for the Derivative of x^sinx</h2>
67
<p>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.</p>
66
<p>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.</p>
68
<p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
67
<p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
69
<p>▶</p>
68
<p>▶</p>
70
<h2>Jaskaran Singh Saluja</h2>
69
<h2>Jaskaran Singh Saluja</h2>
71
<h3>About the Author</h3>
70
<h3>About the Author</h3>
72
<p>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.</p>
71
<p>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.</p>
73
<h3>Fun Fact</h3>
72
<h3>Fun Fact</h3>
74
<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>
73
<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>