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2026-01-01
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<p>Last updated on<strong>September 12, 2025</strong></p>
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<p>Last updated on<strong>September 12, 2025</strong></p>
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<p>We use the derivative of a negative exponent function, which is a key concept in calculus, to understand how functions with negative powers change in response to slight changes in x. Derivatives are vital for calculating profit or loss and understanding rates of change in real-life situations. We will now discuss the derivative of negative exponent functions in detail.</p>
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<p>We use the derivative of a negative exponent function, which is a key concept in calculus, to understand how functions with negative powers change in response to slight changes in x. Derivatives are vital for calculating profit or loss and understanding rates of change in real-life situations. We will now discuss the derivative of negative exponent functions in detail.</p>
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<h2>What is the Derivative of a Negative Exponent Function?</h2>
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<h2>What is the Derivative of a Negative Exponent Function?</h2>
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<p>We now understand the derivative of a<a>function</a>with a<a>negative exponent</a>. It is commonly represented as d/dx (x⁻ⁿ) or (x⁻ⁿ)', and its value is -n·x⁻ⁿ⁻¹.</p>
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<p>We now understand the derivative of a<a>function</a>with a<a>negative exponent</a>. It is commonly represented as d/dx (x⁻ⁿ) or (x⁻ⁿ)', and its value is -n·x⁻ⁿ⁻¹.</p>
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<p>Functions with negative exponents have clearly defined derivatives, indicating they are differentiable within their domain. The key concepts are mentioned below:</p>
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<p>Functions with negative exponents have clearly defined derivatives, indicating they are differentiable within their domain. The key concepts are mentioned below:</p>
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<p>Power Rule: Rule for differentiating functions of the form xⁿ, applicable to x⁻ⁿ as well.</p>
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<p>Power Rule: Rule for differentiating functions of the form xⁿ, applicable to x⁻ⁿ as well.</p>
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<p>Negative Exponent: x⁻ⁿ = 1/xⁿ, an important identity to understand.</p>
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<p>Negative Exponent: x⁻ⁿ = 1/xⁿ, an important identity to understand.</p>
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<h2>Derivative of a Negative Exponent Formula</h2>
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<h2>Derivative of a Negative Exponent Formula</h2>
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<p>The derivative of x⁻ⁿ can be denoted as d/dx (x⁻ⁿ) or (x⁻ⁿ)'.</p>
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<p>The derivative of x⁻ⁿ can be denoted as d/dx (x⁻ⁿ) or (x⁻ⁿ)'.</p>
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<p>The<a>formula</a>we use to differentiate x⁻ⁿ is: d/dx (x⁻ⁿ) = -n·x⁻ⁿ⁻¹ (or) (x⁻ⁿ)' = -n·x⁻ⁿ⁻¹</p>
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<p>The<a>formula</a>we use to differentiate x⁻ⁿ is: d/dx (x⁻ⁿ) = -n·x⁻ⁿ⁻¹ (or) (x⁻ⁿ)' = -n·x⁻ⁿ⁻¹</p>
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<p>The formula applies to all x where x ≠ 0.</p>
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<p>The formula applies to all x where x ≠ 0.</p>
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<h2>Proofs of the Derivative of a Negative Exponent</h2>
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<h2>Proofs of the Derivative of a Negative Exponent</h2>
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<p>We can derive the derivative of x⁻ⁿ using proofs. To show this, we will use algebraic manipulation along with the rules of differentiation. There are several methods we use to prove this, such as:</p>
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<p>We can derive the derivative of x⁻ⁿ using proofs. To show this, we will use algebraic manipulation along with the rules of differentiation. There are several methods we use to prove this, such as:</p>
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<ol><li>By First Principle</li>
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<ol><li>By First Principle</li>
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<li>Using Power Rule</li>
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<li>Using Power Rule</li>
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<li>Using Negative Exponent</li>
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<li>Using Negative Exponent</li>
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</ol><p>Identity We will now demonstrate that the differentiation of x⁻ⁿ results in -n·x⁻ⁿ⁻¹ using the above-mentioned methods:</p>
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</ol><p>Identity We will now demonstrate that the differentiation of x⁻ⁿ results in -n·x⁻ⁿ⁻¹ using the above-mentioned methods:</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⁻ⁿ can 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⁻ⁿ can 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⁻ⁿ using the first principle, we will consider f(x) = x⁻ⁿ. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>To find the derivative of x⁻ⁿ using the first principle, we will consider f(x) = x⁻ⁿ. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>Given that f(x) = x⁻ⁿ, we write f(x + h) = (x + h)⁻ⁿ.</p>
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<p>Given that f(x) = x⁻ⁿ, we write f(x + h) = (x + h)⁻ⁿ.</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [(x + h)⁻ⁿ - x⁻ⁿ] / h</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [(x + h)⁻ⁿ - x⁻ⁿ] / h</p>
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<p>Using<a>binomial</a>expansion and simplifying, f'(x) = -n·x⁻ⁿ⁻¹ Hence, proved.</p>
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<p>Using<a>binomial</a>expansion and simplifying, f'(x) = -n·x⁻ⁿ⁻¹ Hence, proved.</p>
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<h3>Using Power Rule</h3>
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<h3>Using Power Rule</h3>
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<p>To prove the differentiation of x⁻ⁿ using the<a>power</a>rule, We use the formula: d/dx (xⁿ) = n·xⁿ⁻¹ Set n to be negative, i.e., n = -m, d/dx (x⁻ⁿ) = -n·x⁻ⁿ⁻¹</p>
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<p>To prove the differentiation of x⁻ⁿ using the<a>power</a>rule, We use the formula: d/dx (xⁿ) = n·xⁿ⁻¹ Set n to be negative, i.e., n = -m, d/dx (x⁻ⁿ) = -n·x⁻ⁿ⁻¹</p>
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<p>Thus, the derivative of x⁻ⁿ is -n·x⁻ⁿ⁻¹.</p>
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<p>Thus, the derivative of x⁻ⁿ is -n·x⁻ⁿ⁻¹.</p>
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<h3>Using Negative Exponent</h3>
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<h3>Using Negative Exponent</h3>
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<p>Identity Consider x⁻ⁿ = 1/xⁿ Differentiate using the quotient rule, d/dx (1/xⁿ) = -n·x⁻ⁿ⁻¹</p>
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<p>Identity Consider x⁻ⁿ = 1/xⁿ Differentiate using the quotient rule, d/dx (1/xⁿ) = -n·x⁻ⁿ⁻¹</p>
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<p>Hence, the derivative of x⁻ⁿ is -n·x⁻ⁿ⁻¹.</p>
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<p>Hence, the derivative of x⁻ⁿ is -n·x⁻ⁿ⁻¹.</p>
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<h2>Higher-Order Derivatives of Negative Exponent Functions</h2>
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<h2>Higher-Order Derivatives of Negative Exponent Functions</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 bit challenging.</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 bit challenging.</p>
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<p>To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like x⁻ⁿ.</p>
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<p>To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like x⁻ⁿ.</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 x⁻ⁿ, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change (continuing for higher-order derivatives).</p>
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<p>For the nth Derivative of x⁻ⁿ, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate 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 is 0, the derivative is undefined because x⁻ⁿ involves<a>division by zero</a>at this point.</p>
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<p>When x is 0, the derivative is undefined because x⁻ⁿ involves<a>division by zero</a>at this point.</p>
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<p>When n = 1, the derivative of x⁻¹ = -1·x⁻², which is -1/x².</p>
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<p>When n = 1, the derivative of x⁻¹ = -1·x⁻², which is -1/x².</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Negative Exponents</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Negative Exponents</h2>
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<p>Students frequently make mistakes when differentiating functions with negative exponents. 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 functions with negative exponents. 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⁻² · x⁻³)</p>
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<p>Calculate the derivative of (x⁻² · 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>Here, we have f(x) = x⁻² · x⁻³. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x⁻² and v = x⁻³. Let’s differentiate each term, u′ = d/dx (x⁻²) = -2x⁻³ v′ = d/dx (x⁻³) = -3x⁻⁴</p>
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<p>Here, we have f(x) = x⁻² · x⁻³. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x⁻² and v = x⁻³. Let’s differentiate each term, u′ = d/dx (x⁻²) = -2x⁻³ v′ = d/dx (x⁻³) = -3x⁻⁴</p>
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<p>Substituting into the given equation, f'(x) = (-2x⁻³)(x⁻³) + (x⁻²)(-3x⁻⁴)</p>
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<p>Substituting into the given equation, f'(x) = (-2x⁻³)(x⁻³) + (x⁻²)(-3x⁻⁴)</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = -2x⁻⁶ - 3x⁻⁶ f'(x) = -5x⁻⁶</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = -2x⁻⁶ - 3x⁻⁶ f'(x) = -5x⁻⁶</p>
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<p>Thus, the derivative of the specified function is -5x⁻⁶.</p>
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<p>Thus, the derivative of the specified function is -5x⁻⁶.</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 lamp post casts a shadow represented by the function y = x⁻¹, where y represents the length of the shadow at a distance x from the lamp post. If x = 2 meters, measure the rate of change of the shadow length.</p>
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<p>A lamp post casts a shadow represented by the function y = x⁻¹, where y represents the length of the shadow at a distance x from the lamp post. If x = 2 meters, measure the rate of change of the shadow length.</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⁻¹ (length of the shadow)...(1)</p>
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<p>We have y = x⁻¹ (length of the shadow)...(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⁻¹: dy/dx = -1·x⁻² Given x = 2 (substitute this into the derivative)</p>
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<p>Take the derivative x⁻¹: dy/dx = -1·x⁻² Given x = 2 (substitute this into the derivative)</p>
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<p>dy/dx = -1·2⁻² dy/dx = -1/4</p>
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<p>dy/dx = -1·2⁻² dy/dx = -1/4</p>
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<p>Hence, we get the rate of change of the shadow length at a distance x = 2 as -1/4.</p>
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<p>Hence, we get the rate of change of the shadow length at a distance x = 2 as -1/4.</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 shadow length at x = 2 as -1/4, which means that at a given point, the length of the shadow decreases at a rate of 1/4 units per unit increase in distance from the lamp post.</p>
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<p>We find the rate of change of the shadow length at x = 2 as -1/4, which means that at a given point, the length of the shadow decreases at a rate of 1/4 units per unit increase in distance from the lamp post.</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⁻².</p>
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<p>Derive the second derivative of the function y = 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>The first step is to find the first derivative, dy/dx = -2x⁻³...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = -2x⁻³...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-2x⁻³]</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-2x⁻³]</p>
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<p>Here we use the power rule, d²y/dx² = -2(-3)x⁻⁴ d²y/dx² = 6x⁻⁴</p>
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<p>Here we use the power rule, d²y/dx² = -2(-3)x⁻⁴ d²y/dx² = 6x⁻⁴</p>
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<p>Therefore, the second derivative of the function y = x⁻² is 6x⁻⁴.</p>
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<p>Therefore, the second derivative of the function y = x⁻² is 6x⁻⁴.</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 -2x⁻³. We then simplify the terms to find the final answer.</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 -2x⁻³. We then simplify the terms to find the final answer.</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⁻¹)²) = -2x⁻³.</p>
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<p>Prove: d/dx ((x⁻¹)²) = -2x⁻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>Let’s start using the chain rule: Consider y = (x⁻¹)²</p>
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<p>Let’s start using the chain rule: Consider y = (x⁻¹)²</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2(x⁻¹)·d/dx [x⁻¹]</p>
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<p>To differentiate, we use the chain rule: dy/dx = 2(x⁻¹)·d/dx [x⁻¹]</p>
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<p>Since the derivative of x⁻¹ is -1·x⁻², dy/dx = 2(x⁻¹)(-1·x⁻²)</p>
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<p>Since the derivative of x⁻¹ is -1·x⁻², dy/dx = 2(x⁻¹)(-1·x⁻²)</p>
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<p>dy/dx = -2x⁻³ Hence proved.</p>
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<p>dy/dx = -2x⁻³ 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⁻¹ 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⁻¹ 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⁻²/x)</p>
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<p>Solve: d/dx (x⁻²/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⁻²/x) = (d/dx (x⁻²)·x - x⁻²·d/dx(x))/x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (x⁻²/x) = (d/dx (x⁻²)·x - x⁻²·d/dx(x))/x²</p>
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<p>We will substitute d/dx (x⁻²) = -2x⁻³ and d/dx(x) = 1 = (-2x⁻³·x - x⁻²·1)/x² = (-2x⁻² - x⁻²)/x² = -3x⁻²/x² = -3x⁻⁴</p>
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<p>We will substitute d/dx (x⁻²) = -2x⁻³ and d/dx(x) = 1 = (-2x⁻³·x - x⁻²·1)/x² = (-2x⁻² - x⁻²)/x² = -3x⁻²/x² = -3x⁻⁴</p>
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<p>Therefore, d/dx (x⁻²/x) = -3x⁻⁴.</p>
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<p>Therefore, d/dx (x⁻²/x) = -3x⁻⁴.</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 Negative Exponents</h2>
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<h2>FAQs on the Derivative of Negative Exponents</h2>
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<h3>1.Find the derivative of x⁻².</h3>
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<h3>1.Find the derivative of x⁻².</h3>
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<p>Using the power rule for negative<a>exponents</a>, d/dx (x⁻²) = -2x⁻³.</p>
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<p>Using the power rule for negative<a>exponents</a>, d/dx (x⁻²) = -2x⁻³.</p>
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<h3>2.Can we use the derivative of negative exponents in real life?</h3>
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<h3>2.Can we use the derivative of negative exponents in real life?</h3>
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<p>Yes, we can use the derivative of negative exponents in real life to calculate rates of change, especially in contexts involving inversely proportional relationships.</p>
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<p>Yes, we can use the derivative of negative exponents in real life to calculate rates of change, especially in contexts involving inversely proportional relationships.</p>
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<h3>3.Is it possible to take the derivative of x⁻ⁿ at the point where x = 0?</h3>
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<h3>3.Is it possible to take the derivative of x⁻ⁿ at the point where x = 0?</h3>
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<p>No, x = 0 is a point where x⁻ⁿ 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⁻ⁿ 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⁻²/x?</h3>
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<h3>4.What rule is used to differentiate x⁻²/x?</h3>
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<p>We use the quotient rule to differentiate x⁻²/x, d/dx (x⁻²/x) = (-2x⁻³·x - x⁻²·1)/x² = -3x⁻⁴.</p>
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<p>We use the quotient rule to differentiate x⁻²/x, d/dx (x⁻²/x) = (-2x⁻³·x - x⁻²·1)/x² = -3x⁻⁴.</p>
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<h3>5.Are the derivatives of x⁻² and x² the same?</h3>
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<h3>5.Are the derivatives of x⁻² and x² the same?</h3>
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<p>No, they are different. The derivative of x⁻² is -2x⁻³, while the derivative of x² is 2x.</p>
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<p>No, they are different. The derivative of x⁻² is -2x⁻³, while the derivative of x² is 2x.</p>
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<h2>Important Glossaries for the Derivative of Negative Exponents</h2>
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<h2>Important Glossaries for the Derivative of Negative Exponents</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>Negative Exponent:</strong>A term of the form x⁻ⁿ, which is equal to 1/xⁿ.</li>
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</ul><ul><li><strong>Negative Exponent:</strong>A term of the form x⁻ⁿ, which is equal to 1/xⁿ.</li>
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</ul><ul><li><strong>Power Rule:</strong>A basic rule in calculus used to differentiate functions of the form xⁿ.</li>
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</ul><ul><li><strong>Power Rule:</strong>A basic rule in calculus used to differentiate functions of the form xⁿ.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to differentiate compositions of functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule used to differentiate compositions of functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate functions that are ratios of other functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule used to differentiate functions that are ratios of other functions.</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>