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2026-01-01
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<p>Last updated on<strong>September 22, 2025</strong></p>
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<p>Last updated on<strong>September 22, 2025</strong></p>
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<p>We use the derivative of x^-3, which is -3x^-4, to understand how the function changes in response to a slight change in x. Derivatives play a crucial role in various real-life applications, such as calculating rates of change. We will now discuss the derivative of x^-3 in detail.</p>
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<p>We use the derivative of x^-3, which is -3x^-4, to understand how the function changes in response to a slight change in x. Derivatives play a crucial role in various real-life applications, such as calculating rates of change. We will now discuss the derivative of x^-3 in detail.</p>
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<h2>What is the Derivative of x^-3?</h2>
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<h2>What is the Derivative of x^-3?</h2>
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<p>We now understand the derivative of x^-3. It is commonly represented as d/dx (x^-3) or (x^-3)', and its value is -3x^-4. The<a>function</a>x^-3 has a well-defined derivative, indicating it is differentiable across its domain.</p>
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<p>We now understand the derivative of x^-3. It is commonly represented as d/dx (x^-3) or (x^-3)', and its value is -3x^-4. The<a>function</a>x^-3 has a well-defined derivative, indicating it is differentiable across its domain.</p>
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<p>The key concepts are mentioned below:</p>
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<p>The key concepts are mentioned below:</p>
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<p><strong>Power Rule:</strong>A fundamental rule for differentiating functions of the form x^n.</p>
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<p><strong>Power Rule:</strong>A fundamental rule for differentiating functions of the form x^n.</p>
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<p><strong>Negative Exponents:</strong>Understanding how to differentiate functions with negative<a>powers</a>.</p>
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<p><strong>Negative Exponents:</strong>Understanding how to differentiate functions with negative<a>powers</a>.</p>
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<h2>Derivative of x^-3 Formula</h2>
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<h2>Derivative of x^-3 Formula</h2>
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<p>The derivative of x^-3 can be denoted as d/dx (x^-3) or (x^-3)'.</p>
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<p>The derivative of x^-3 can be denoted as d/dx (x^-3) or (x^-3)'.</p>
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<p>The<a>formula</a>we use to differentiate x^-3 is: d/dx (x^-3) = -3x^-4</p>
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<p>The<a>formula</a>we use to differentiate x^-3 is: d/dx (x^-3) = -3x^-4</p>
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<p>The formula applies to all x except where x = 0, as<a>division by zero</a>is undefined.</p>
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<p>The formula applies to all x except where x = 0, as<a>division by zero</a>is undefined.</p>
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<h2>Proofs of the Derivative of x^-3</h2>
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<h2>Proofs of the Derivative of x^-3</h2>
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<p>We can derive the derivative of x^-3 using proofs. To show this, we will use differentiation rules. There are several methods we use to prove this, such as:</p>
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<p>We can derive the derivative of x^-3 using proofs. To show this, we will use differentiation rules. There are several methods we use to prove this, such as:</p>
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<ol><li>By Power Rule</li>
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<ol><li>By Power Rule</li>
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<li>By First Principle</li>
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<li>By First Principle</li>
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</ol><h3>By Power Rule</h3>
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</ol><h3>By Power Rule</h3>
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<p>The derivative of x^-3 can be easily found using the power rule, which states that d/dx (x^n) = n*x^(n-1). For x^-3, n = -3. Thus, d/dx (x^-3) = -3*x^(-3-1) = -3x^-4.</p>
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<p>The derivative of x^-3 can be easily found using the power rule, which states that d/dx (x^n) = n*x^(n-1). For x^-3, n = -3. Thus, d/dx (x^-3) = -3*x^(-3-1) = -3x^-4.</p>
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<h3>By First Principle</h3>
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<h3>By First Principle</h3>
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<p>The derivative of x^-3 can also be proved using the first principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>The derivative of x^-3 can also be proved using the first principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>To find the derivative of x^-3 using the first principle, we consider f(x) = x^-3. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h</p>
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<p>To find the derivative of x^-3 using the first principle, we consider f(x) = x^-3. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h</p>
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<p>Given that f(x) = x^-3, we write f(x + h) = (x + h)^-3.</p>
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<p>Given that f(x) = x^-3, we write f(x + h) = (x + h)^-3.</p>
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<p>Substituting these into the<a>equation</a>, f'(x) = limₕ→₀ [(x + h)^-3 - x^-3] / h</p>
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<p>Substituting these into the<a>equation</a>, f'(x) = limₕ→₀ [(x + h)^-3 - x^-3] / h</p>
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<p>Simplifying using<a>binomial</a>expansion and limits, f'(x) = -3x^-4.</p>
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<p>Simplifying using<a>binomial</a>expansion and limits, f'(x) = -3x^-4.</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<h2>Higher-Order Derivatives of x^-3</h2>
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<h2>Higher-Order Derivatives of x^-3</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 offer insights into the behavior<a>of functions</a>.</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 offer insights into the behavior<a>of functions</a>.</p>
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<p>For the first derivative of a function, we write f′(x), indicating how the function changes or its slope at a particular point. The second derivative is derived from the first derivative, 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), indicating how the function changes or its slope at a particular point. The second derivative is derived from the first derivative, 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 x^-3, we generally use f^(n)(x) for the nth derivative of a function f(x), which tells us the change in the<a>rate</a>of change, continuing for higher-order derivatives.</p>
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<p>For the nth derivative of x^-3, we generally use f^(n)(x) for the nth derivative of a function f(x), which tells us the change in the<a>rate</a>of change, continuing for higher-order derivatives.</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>When x = 0, the derivative is undefined because x^-3 is not defined at zero. When x = 1, the derivative of x^-3 = -3*1^-4 = -3.</p>
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<p>When x = 0, the derivative is undefined because x^-3 is not defined at zero. When x = 1, the derivative of x^-3 = -3*1^-4 = -3.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of x^-3</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of x^-3</h2>
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<p>Students frequently make mistakes when differentiating x^-3. 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 x^-3. 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 (x^-3 * x^2)</p>
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<p>Calculate the derivative of (x^-3 * x^2)</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) = x^-3 * x^2.</p>
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<p>Here, we have f(x) = x^-3 * x^2.</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x^-3 and v = x^2.</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x^-3 and v = x^2.</p>
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<p>Let’s differentiate each term, u′= d/dx (x^-3) = -3x^-4 v′= d/dx (x^2) = 2x</p>
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<p>Let’s differentiate each term, u′= d/dx (x^-3) = -3x^-4 v′= d/dx (x^2) = 2x</p>
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<p>Substituting into the given equation, f'(x) = (-3x^-4) * (x^2) + (x^-3) * (2x) = -3x^-2 + 2x^-2</p>
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<p>Substituting into the given equation, f'(x) = (-3x^-4) * (x^2) + (x^-3) * (2x) = -3x^-2 + 2x^-2</p>
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<p>Simplifying terms to get the final answer, f'(x) = -x^-2</p>
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<p>Simplifying terms to get the final answer, f'(x) = -x^-2</p>
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<p>Thus, the derivative of the specified function is -x^-2.</p>
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<p>Thus, the derivative of the specified function is -x^-2.</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. 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>We find the derivative of the given function by dividing the function into two parts. 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’s depreciation of a machine is modeled by the function y = x^-3, where y represents the depreciation value at time x. If x = 2 years, calculate the rate of depreciation.</p>
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<p>A company’s depreciation of a machine is modeled by the function y = x^-3, where y represents the depreciation value at time x. If x = 2 years, calculate the rate of depreciation.</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 y = x^-3 (depreciation model)...(1)</p>
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<p>We have y = x^-3 (depreciation model)...(1)</p>
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<p>Now, we will differentiate the equation (1)</p>
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<p>Now, we will differentiate the equation (1)</p>
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<p>Take the derivative x^-3: dy/dx = -3x^-4 Given x = 2 (substitute this into the derivative) dy/dx = -3*(2^-4) = -3/16</p>
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<p>Take the derivative x^-3: dy/dx = -3x^-4 Given x = 2 (substitute this into the derivative) dy/dx = -3*(2^-4) = -3/16</p>
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<p>Hence, we get the rate of depreciation at a time x = 2 years as -3/16.</p>
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<p>Hence, we get the rate of depreciation at a time x = 2 years as -3/16.</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 depreciation at x = 2 as -3/16, which indicates that the depreciation value decreases at this rate at the given point in time.</p>
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<p>We find the rate of depreciation at x = 2 as -3/16, which indicates that the depreciation value decreases at this rate at the given point in time.</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^-3.</p>
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<p>Derive the second derivative of the function y = x^-3.</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 = -3x^-4...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = -3x^-4...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-3x^-4] d²y/dx² = 12x^-5</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-3x^-4] d²y/dx² = 12x^-5</p>
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<p>Therefore, the second derivative of the function y = x^-3 is 12x^-5.</p>
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<p>Therefore, the second derivative of the function y = x^-3 is 12x^-5.</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. Using the power rule, we differentiate -3x^-4 to find the second derivative: 12x^-5.</p>
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<p>We use the step-by-step process, where we start with the first derivative. Using the power rule, we differentiate -3x^-4 to find the second derivative: 12x^-5.</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)^2) = -6x^-7.</p>
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<p>Prove: d/dx ((x^-3)^2) = -6x^-7.</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 chain rule: Consider y = (x^-3)^2 = (x^-3)^2</p>
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<p>Let’s start using the chain rule: Consider y = (x^-3)^2 = (x^-3)^2</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2(x^-3) * d/dx [x^-3]</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2(x^-3) * d/dx [x^-3]</p>
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<p>Since the derivative of x^-3 is -3x^-4, dy/dx = 2(x^-3) * (-3x^-4) = -6x^-7</p>
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<p>Since the derivative of x^-3 is -3x^-4, dy/dx = 2(x^-3) * (-3x^-4) = -6x^-7</p>
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<p>Hence proved.</p>
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<p>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 chain rule to differentiate the equation. Then, we replace x^-3 with its derivative. As a final step, we simplify to derive the equation.</p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace x^-3 with its derivative. As a final step, we simplify to derive the 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 (x^-3/x)</p>
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<p>Solve: d/dx (x^-3/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 quotient rule: d/dx (x^-3/x) = (d/dx (x^-3) * x - x^-3 * d/dx(x))/x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (x^-3/x) = (d/dx (x^-3) * x - x^-3 * d/dx(x))/x²</p>
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<p>We will substitute d/dx (x^-3) = -3x^-4 and d/dx (x) = 1 = ( (-3x^-4) * x - x^-3 * 1) / x² = (-3x^-3 - x^-3) / x² = -4x^-3 / x² = -4x^-5</p>
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<p>We will substitute d/dx (x^-3) = -3x^-4 and d/dx (x) = 1 = ( (-3x^-4) * x - x^-3 * 1) / x² = (-3x^-3 - x^-3) / x² = -4x^-3 / x² = -4x^-5</p>
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<p>Therefore, d/dx (x^-3/x) = -4x^-5.</p>
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<p>Therefore, d/dx (x^-3/x) = -4x^-5.</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 quotient rule. As a final step, we simplify the equation to obtain the final result.</p>
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<p>In this process, we differentiate the given function using the quotient rule. As a final step, we simplify the equation 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 x^-3</h2>
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<h2>FAQs on the Derivative of x^-3</h2>
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<h3>1.Find the derivative of x^-3.</h3>
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<h3>1.Find the derivative of x^-3.</h3>
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<p>Using the power rule for x^-3 gives: d/dx (x^-3) = -3x^-4.</p>
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<p>Using the power rule for x^-3 gives: d/dx (x^-3) = -3x^-4.</p>
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<h3>2.Can we use the derivative of x^-3 in real life?</h3>
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<h3>2.Can we use the derivative of x^-3 in real life?</h3>
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<p>Yes, the derivative of x^-3 can be used in real life to model situations where the rate of change is important, such as in physics or economics.</p>
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<p>Yes, the derivative of x^-3 can be used in real life to model situations where the rate of change is important, such as in physics or economics.</p>
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<h3>3.Is it possible to take the derivative of x^-3 at the point where x = 0?</h3>
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<h3>3.Is it possible to take the derivative of x^-3 at the point where x = 0?</h3>
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<p>No, x = 0 is a point where x^-3 is undefined, so it is impossible to take the derivative at this point (since the function does not exist there).</p>
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<p>No, x = 0 is a point where x^-3 is undefined, so it is impossible to take the derivative at this point (since the function does not exist there).</p>
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<h3>4.What rule is used to differentiate x^-3/x?</h3>
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<h3>4.What rule is used to differentiate x^-3/x?</h3>
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<p>We use the quotient rule to differentiate x^-3/x, d/dx (x^-3/x) = (x * (-3x^-4) - x^-3 * 1) / x².</p>
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<p>We use the quotient rule to differentiate x^-3/x, d/dx (x^-3/x) = (x * (-3x^-4) - x^-3 * 1) / x².</p>
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<h3>5.Are the derivatives of x^-3 and (x^3)^-1 the same?</h3>
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<h3>5.Are the derivatives of x^-3 and (x^3)^-1 the same?</h3>
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<p>Yes, they are the same because (x^3)^-1 is equivalent to x^-3, and their derivative is -3x^-4.</p>
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<p>Yes, they are the same because (x^3)^-1 is equivalent to x^-3, and their derivative is -3x^-4.</p>
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<h2>Important Glossaries for the Derivative of x^-3</h2>
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<h2>Important Glossaries for the Derivative of x^-3</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in x.</li>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in x.</li>
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</ul><ul><li><strong>Power Rule:</strong>A fundamental rule used to differentiate functions of the form x^n.</li>
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</ul><ul><li><strong>Power Rule:</strong>A fundamental rule used to differentiate functions of the form x^n.</li>
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</ul><ul><li><strong>Negative Exponent:</strong>A concept in mathematics where a negative exponent indicates the reciprocal of the positive exponent.</li>
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</ul><ul><li><strong>Negative Exponent:</strong>A concept in mathematics where a negative exponent indicates the reciprocal of the positive exponent.</li>
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</ul><ul><li><strong>First Derivative:</strong>The initial result of differentiating a function, giving the rate of change of the function.</li>
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</ul><ul><li><strong>First Derivative:</strong>The initial result of differentiating a function, giving the rate of change of the function.</li>
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</ul><ul><li><strong>Undefined Point:</strong>A point at which a function is not defined, often due to division by zero or other mathematical restrictions.</li>
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</ul><ul><li><strong>Undefined Point:</strong>A point at which a function is not defined, often due to division by zero or other mathematical restrictions.</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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<p>▶</p>
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<p>▶</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>