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2026-01-01
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<p>Last updated on<strong>September 27, 2025</strong></p>
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<p>Last updated on<strong>September 27, 2025</strong></p>
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<p>We use the derivative of multiplication to measure how the product of two functions 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 multiplication in detail.</p>
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<p>We use the derivative of multiplication to measure how the product of two functions 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 multiplication in detail.</p>
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<h2>What is the Derivative of Multiplication?</h2>
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<h2>What is the Derivative of Multiplication?</h2>
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<p>To understand the derivative<a>of</a><a>multiplication</a>, we use the<a>product</a>rule. It is commonly represented as d/dx (u·v), where u and v are<a>functions</a>of x. The derivative is given by u'v + uv'. This rule indicates that the product of two differentiable functions is itself differentiable.</p>
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<p>To understand the derivative<a>of</a><a>multiplication</a>, we use the<a>product</a>rule. It is commonly represented as d/dx (u·v), where u and v are<a>functions</a>of x. The derivative is given by u'v + uv'. This rule indicates that the product of two differentiable functions is itself differentiable.</p>
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<p>The key concepts are mentioned below: Product Rule: A rule for differentiating the product of two functions. Derivative: The<a>rate</a>of change of a function with respect to a<a>variable</a>. Function: A mathematical relation between a<a>set</a>of inputs and outputs, where each input is related to exactly one output.</p>
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<p>The key concepts are mentioned below: Product Rule: A rule for differentiating the product of two functions. Derivative: The<a>rate</a>of change of a function with respect to a<a>variable</a>. Function: A mathematical relation between a<a>set</a>of inputs and outputs, where each input is related to exactly one output.</p>
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<h2>Derivative of Multiplication Formula</h2>
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<h2>Derivative of Multiplication Formula</h2>
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<p>The derivative of multiplication can be denoted as d/dx (u·v). The<a>formula</a>we use to differentiate the product of two functions is: d/dx (u·v) = u'v + uv'</p>
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<p>The derivative of multiplication can be denoted as d/dx (u·v). The<a>formula</a>we use to differentiate the product of two functions is: d/dx (u·v) = u'v + uv'</p>
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<p>This formula applies to all x where both functions u and v are differentiable.</p>
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<p>This formula applies to all x where both functions u and v are differentiable.</p>
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<h2>Proofs of the Derivative of Multiplication</h2>
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<h2>Proofs of the Derivative of Multiplication</h2>
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<p>We can derive the derivative of multiplication using proofs. To show this, we will use the rules of differentiation. Here are several methods we use to prove this:</p>
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<p>We can derive the derivative of multiplication using proofs. To show this, we will use the rules of differentiation. Here are several methods we use to prove this:</p>
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<h2><strong>Using the Product Rule</strong></h2>
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<h2><strong>Using the Product Rule</strong></h2>
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<p>According to the product rule for differentiation, if u(x) and v(x) are differentiable functions, then the derivative of their product is given by: d/dx (u·v) = u'v + uv' This formula comes from considering the limit of the difference<a>quotient</a>of the product of two functions.</p>
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<p>According to the product rule for differentiation, if u(x) and v(x) are differentiable functions, then the derivative of their product is given by: d/dx (u·v) = u'v + uv' This formula comes from considering the limit of the difference<a>quotient</a>of the product of two functions.</p>
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<h2><strong>Using the First Principle</strong></h2>
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<h2><strong>Using the First Principle</strong></h2>
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<p>The derivative of the product u·v can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. Consider f(x) = u(x)·v(x). Its derivative can be expressed as the following limit: f'(x) = limₕ→₀ [(u(x + h)v(x + h) - u(x)v(x))] / h We can expand u(x + h) and v(x + h) using their respective derivatives, and upon simplification, the limit resolves to: f'(x) = u'(x)v(x) + u(x)v'(x) Thus, the derivative of the product is the<a>sum</a>of each function's derivative times the other function.</p>
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<p>The derivative of the product u·v can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. Consider f(x) = u(x)·v(x). Its derivative can be expressed as the following limit: f'(x) = limₕ→₀ [(u(x + h)v(x + h) - u(x)v(x))] / h We can expand u(x + h) and v(x + h) using their respective derivatives, and upon simplification, the limit resolves to: f'(x) = u'(x)v(x) + u(x)v'(x) Thus, the derivative of the product is the<a>sum</a>of each function's derivative times the other function.</p>
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<h2>Higher-Order Derivatives of Multiplication</h2>
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<h2>Higher-Order Derivatives of Multiplication</h2>
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<p>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 multiplication of<a>multiple</a>functions.</p>
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<p>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 multiplication of<a>multiple</a>functions.</p>
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<p>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.</p>
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<p>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.</p>
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<p>For the nth Derivative of a product of functions, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change.</p>
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<p>For the nth Derivative of a product of functions, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change.</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>When one of the functions is a<a>constant</a>, the derivative of the product simplifies to the derivative of the non-constant function multiplied by the constant.</p>
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<p>When one of the functions is a<a>constant</a>, the derivative of the product simplifies to the derivative of the non-constant function multiplied by the constant.</p>
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<p>When one of the functions is zero, the derivative is zero, as the product itself is zero.</p>
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<p>When one of the functions is zero, the derivative is zero, as the product itself is zero.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Multiplication</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Multiplication</h2>
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<p>Students frequently make mistakes when differentiating products of functions. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:</p>
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<p>Students frequently make mistakes when differentiating products of functions. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:</p>
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<h3>Problem 1</h3>
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<h3>Problem 1</h3>
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<p>Calculate the derivative of (x²·e^x)</p>
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<p>Calculate the derivative of (x²·e^x)</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Here, we have f(x) = x²·e^x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x² and v = e^x. Let’s differentiate each term, u′ = d/dx (x²) = 2x v′ = d/dx (e^x) = e^x substituting into the given equation, f'(x) = (2x)·(e^x) + (x²)·(e^x) Let’s simplify terms to get the final answer, f'(x) = 2xe^x + x²e^x Thus, the derivative of the specified function is 2xe^x + x²e^x.</p>
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<p>Here, we have f(x) = x²·e^x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x² and v = e^x. Let’s differentiate each term, u′ = d/dx (x²) = 2x v′ = d/dx (e^x) = e^x substituting into the given equation, f'(x) = (2x)·(e^x) + (x²)·(e^x) Let’s simplify terms to get the final answer, f'(x) = 2xe^x + x²e^x Thus, the derivative of the specified function is 2xe^x + x²e^x.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the derivative of the given function by dividing the function into two parts.</p>
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<p>We find the derivative of the given function by dividing the function into two parts.</p>
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<p>The first step is finding its derivative and then combining them using the product rule to get the final result.</p>
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<p>The first step is finding its derivative and then combining them using the product rule to get the final result.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 2</h3>
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<h3>Problem 2</h3>
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<p>A company produces widgets, and the production cost is represented by the function C(x) = x·ln(x), where x is the number of widgets produced. Calculate the rate of change of the cost when x = 10.</p>
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<p>A company produces widgets, and the production cost is represented by the function C(x) = x·ln(x), where x is the number of widgets produced. Calculate the rate of change of the cost when x = 10.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>We have C(x) = x·ln(x). Now, we will differentiate the equation using the product rule, dC/dx = u′v + uv′ Here, u = x and v = ln(x). Let’s differentiate each term, u′ = d/dx (x) = 1 v′ = d/dx (ln(x)) = 1/x substituting into the equation, dC/dx = (1)·(ln(x)) + (x)·(1/x) dC/dx = ln(x) + 1 Given x = 10, substitute this into our derivative, dC/dx = ln(10) + 1 Thus, the rate of change of the cost when producing 10 widgets is ln(10) + 1.</p>
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<p>We have C(x) = x·ln(x). Now, we will differentiate the equation using the product rule, dC/dx = u′v + uv′ Here, u = x and v = ln(x). Let’s differentiate each term, u′ = d/dx (x) = 1 v′ = d/dx (ln(x)) = 1/x substituting into the equation, dC/dx = (1)·(ln(x)) + (x)·(1/x) dC/dx = ln(x) + 1 Given x = 10, substitute this into our derivative, dC/dx = ln(10) + 1 Thus, the rate of change of the cost when producing 10 widgets is ln(10) + 1.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the rate of change of the production cost at x = 10 by differentiating the cost function using the product rule.</p>
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<p>We find the rate of change of the production cost at x = 10 by differentiating the cost function using the product rule.</p>
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<p>We then substitute x = 10 into our derivative to calculate the rate of change.</p>
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<p>We then substitute x = 10 into our derivative to calculate the rate of change.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 3</h3>
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<h3>Problem 3</h3>
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<p>Derive the second derivative of the function y = x·sin(x).</p>
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<p>Derive the second derivative of the function y = x·sin(x).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>The first step is to find the first derivative, dy/dx = cos(x) + x·cos(x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [cos(x) + x·cos(x)] Here we use the product rule, d²y/dx² = -sin(x) + (cos(x) - x·sin(x)) = -sin(x) + cos(x) - x·sin(x) Therefore, the second derivative of the function y = x·sin(x) is -sin(x) + cos(x) - x·sin(x).</p>
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<p>The first step is to find the first derivative, dy/dx = cos(x) + x·cos(x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [cos(x) + x·cos(x)] Here we use the product rule, d²y/dx² = -sin(x) + (cos(x) - x·sin(x)) = -sin(x) + cos(x) - x·sin(x) Therefore, the second derivative of the function y = x·sin(x) is -sin(x) + cos(x) - x·sin(x).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We use the step-by-step process, where we start with the first derivative.</p>
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<p>We use the step-by-step process, where we start with the first derivative.</p>
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<p>Using the product rule, we differentiate the terms and simplify to find the second derivative.</p>
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<p>Using the product rule, we differentiate the terms and simplify to find the second derivative.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 4</h3>
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<h3>Problem 4</h3>
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<p>Prove: d/dx (x^3·ln(x)) = 3x²ln(x) + x².</p>
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<p>Prove: d/dx (x^3·ln(x)) = 3x²ln(x) + x².</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Let’s start using the product rule: Consider y = x^3·ln(x) To differentiate, we use the product rule: dy/dx = u′v + uv′ Where u = x^3 and v = ln(x). Differentiating each, u′ = d/dx (x^3) = 3x² v′ = d/dx (ln(x)) = 1/x Thus, dy/dx = (3x²)·(ln(x)) + (x^3)·(1/x) = 3x²ln(x) + x² Hence proved.</p>
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<p>Let’s start using the product rule: Consider y = x^3·ln(x) To differentiate, we use the product rule: dy/dx = u′v + uv′ Where u = x^3 and v = ln(x). Differentiating each, u′ = d/dx (x^3) = 3x² v′ = d/dx (ln(x)) = 1/x Thus, dy/dx = (3x²)·(ln(x)) + (x^3)·(1/x) = 3x²ln(x) + x² Hence proved.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this step-by-step process, we used the product rule to differentiate the equation.</p>
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<p>In this step-by-step process, we used the product rule to differentiate the equation.</p>
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<p>We then simplify to derive the final equation.</p>
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<p>We then simplify to derive the final equation.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 5</h3>
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<h3>Problem 5</h3>
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<p>Solve: d/dx ((x² + 1)·tan(x))</p>
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<p>Solve: d/dx ((x² + 1)·tan(x))</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>To differentiate the function, we use the product rule: d/dx ((x² + 1)·tan(x)) = (d/dx (x² + 1))·tan(x) + (x² + 1)·d/dx (tan(x)) Differentiating, = (2x)·tan(x) + (x² + 1)·sec²(x) Therefore, d/dx ((x² + 1)·tan(x)) = 2x·tan(x) + (x² + 1)·sec²(x).</p>
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<p>To differentiate the function, we use the product rule: d/dx ((x² + 1)·tan(x)) = (d/dx (x² + 1))·tan(x) + (x² + 1)·d/dx (tan(x)) Differentiating, = (2x)·tan(x) + (x² + 1)·sec²(x) Therefore, d/dx ((x² + 1)·tan(x)) = 2x·tan(x) + (x² + 1)·sec²(x).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this process, we differentiate the given function using the product rule and simplify to obtain the final result.</p>
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<p>In this process, we differentiate the given function using the product rule and simplify to obtain the final result.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h2>FAQs on the Derivative of Multiplication</h2>
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<h2>FAQs on the Derivative of Multiplication</h2>
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<h3>1.What is the derivative of a product of two functions?</h3>
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<h3>1.What is the derivative of a product of two functions?</h3>
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<p>Using the product rule, the derivative of a product of two functions u and v is: d/dx (u·v) = u'v + uv'</p>
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<p>Using the product rule, the derivative of a product of two functions u and v is: d/dx (u·v) = u'v + uv'</p>
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<h3>2.Can the derivative of multiplication be used in real life?</h3>
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<h3>2.Can the derivative of multiplication be used in real life?</h3>
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<p>Yes, the derivative of multiplication can be used in real-life scenarios, such as calculating rates of change in economics, physics, and engineering.</p>
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<p>Yes, the derivative of multiplication can be used in real-life scenarios, such as calculating rates of change in economics, physics, and engineering.</p>
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<h3>3.Is it possible to take the derivative of a product at a point where one function is undefined?</h3>
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<h3>3.Is it possible to take the derivative of a product at a point where one function is undefined?</h3>
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<p>No, the derivative at such a point is undefined if either function is undefined because the product itself is not defined.</p>
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<p>No, the derivative at such a point is undefined if either function is undefined because the product itself is not defined.</p>
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<h3>4.What rule is used to differentiate a product of more than two functions?</h3>
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<h3>4.What rule is used to differentiate a product of more than two functions?</h3>
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<p>For multiple functions, the product rule is extended by applying it iteratively, differentiating one function at a time while keeping others constant.</p>
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<p>For multiple functions, the product rule is extended by applying it iteratively, differentiating one function at a time while keeping others constant.</p>
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<h3>5.Are the derivatives of a product and a quotient the same?</h3>
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<h3>5.Are the derivatives of a product and a quotient the same?</h3>
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<p>No, they are different. The product rule applies to the derivative of a product, while the quotient rule applies to the derivative of a quotient.</p>
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<p>No, they are different. The product rule applies to the derivative of a product, while the quotient rule applies to the derivative of a quotient.</p>
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<h2>Important Glossaries for the Derivative of Multiplication</h2>
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<h2>Important Glossaries for the Derivative of Multiplication</h2>
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<ul><li><strong>Product Rule:</strong>A rule for differentiating the product of two functions.</li>
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<ul><li><strong>Product Rule:</strong>A rule for differentiating the product of two functions.</li>
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</ul><ul><li><strong>Derivative:</strong>The rate of change of a function with respect to a variable.</li>
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</ul><ul><li><strong>Derivative:</strong>The rate of change of a function with respect to a variable.</li>
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</ul><ul><li><strong>Function:</strong>A relation where each input is related to exactly one output.</li>
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</ul><ul><li><strong>Function:</strong>A relation where each input is related to exactly one output.</li>
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</ul><ul><li><strong>Higher-Order Derivative:</strong>Derivatives that are obtained by differentiating a function multiple times.</li>
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</ul><ul><li><strong>Higher-Order Derivative:</strong>Derivatives that are obtained by differentiating a function multiple times.</li>
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</ul><ul><li><strong>Undefined:</strong>A term describing a function or point where a mathematical expression has no meaning.</li>
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</ul><ul><li><strong>Undefined:</strong>A term describing a function or point where a mathematical expression has no meaning.</li>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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<h2>Jaskaran Singh Saluja</h2>
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<h2>Jaskaran Singh Saluja</h2>
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<h3>About the Author</h3>
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<h3>About the Author</h3>
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<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>
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<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>
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<h3>Fun Fact</h3>
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<h3>Fun Fact</h3>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>