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2026-01-01
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<p>Last updated on<strong>September 15, 2025</strong></p>
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<p>Last updated on<strong>September 15, 2025</strong></p>
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<p>The derivative of the function 4√x, represented as d/dx (4√x), is a tool used to measure how the function changes in response to a slight change in x. Derivatives are essential in calculating rates of change, such as speed or economic profit. We will discuss the derivative of 4√x in detail.</p>
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<p>The derivative of the function 4√x, represented as d/dx (4√x), is a tool used to measure how the function changes in response to a slight change in x. Derivatives are essential in calculating rates of change, such as speed or economic profit. We will discuss the derivative of 4√x in detail.</p>
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<h2>What is the Derivative of 4 Square Root of x?</h2>
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<h2>What is the Derivative of 4 Square Root of x?</h2>
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<p>We now understand the derivative<a>of</a>4√x. It is commonly represented as d/dx (4√x) or (4√x)', and its value is 2/x^(1/2). The<a>function</a>4√x has a well-defined derivative, indicating it is differentiable within its domain.</p>
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<p>We now understand the derivative<a>of</a>4√x. It is commonly represented as d/dx (4√x) or (4√x)', and its value is 2/x^(1/2). The<a>function</a>4√x has a well-defined derivative, indicating it is differentiable within 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>Square Root Function:</strong>4√x or 4x^(1/2). </p>
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<p><strong>Square Root Function:</strong>4√x or 4x^(1/2). </p>
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<p><strong>Power Rule:</strong>Rule for differentiating functions of the form x^n. </p>
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<p><strong>Power Rule:</strong>Rule for differentiating functions of the form x^n. </p>
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<p><strong>Chain Rule:</strong>Differentiating composite functions.</p>
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<p><strong>Chain Rule:</strong>Differentiating composite functions.</p>
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<h2>Derivative of 4 Square Root of x Formula</h2>
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<h2>Derivative of 4 Square Root of x Formula</h2>
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<p>The derivative of 4√x can be denoted as d/dx (4√x) or (4√x)'.</p>
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<p>The derivative of 4√x can be denoted as d/dx (4√x) or (4√x)'.</p>
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<p>The<a>formula</a>we use to differentiate 4√x is: d/dx (4√x) = 2/x^(1/2) or (4√x)' = 2/x^(1/2).</p>
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<p>The<a>formula</a>we use to differentiate 4√x is: d/dx (4√x) = 2/x^(1/2) or (4√x)' = 2/x^(1/2).</p>
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<p>This formula applies to all x where x > 0.</p>
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<p>This formula applies to all x where x > 0.</p>
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<h2>Proofs of the Derivative of 4 Square Root of x</h2>
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<h2>Proofs of the Derivative of 4 Square Root of x</h2>
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<p>We can derive the derivative of 4√x 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: - By First Principle - Using Power Rule - Using Chain Rule We will now demonstrate that the differentiation of 4√x results in 2/x^(1/2) using the above-mentioned methods: By First Principle The derivative of 4√x can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of 4√x using the first principle, we will consider f(x) = 4√x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 4√x, we write f(x + h) = 4√(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [4√(x + h) - 4√x] / h We can<a>rationalize</a>the<a>numerator</a>by multiplying by the conjugate: = limₕ→₀ [ (4√(x + h) - 4√x)(4√(x + h) + 4√x) ] / [ h(4√(x + h) + 4√x) ] = limₕ→₀ [ (x + h) - x ] / [ h(4√(x + h) + 4√x) ] = limₕ→₀ h / [ h(4√(x + h) + 4√x) ] = limₕ→₀ 1 / [ 4√(x + h) + 4√x ] = 1 / [ 2√x ] As h approaches 0, 4√(x + h) approaches 4√x, so: f'(x) = 2/x^(1/2). Hence, proved. Using Power Rule To prove the differentiation of 4√x using the<a>power</a>rule, We rewrite the function as 4x^(1/2). By the power rule, d/dx [x^n] = n*x^(n-1), d/dx [4x^(1/2)] = 4 * (1/2) * x^(-1/2) = 2/x^(1/2). Hence, proved. Using Chain Rule We will now prove the derivative of 4√x using the chain rule. The process is as follows: Express the function as a composition: Let u = x^(1/2), then 4√x = 4u. The derivative of u with respect to x is: du/dx = (1/2)x^(-1/2) The derivative of 4u with respect to u is: d/du (4u) = 4 Using the chain rule, d/dx (4u) = d/du (4u) * du/dx = 4 * (1/2)x^(-1/2) = 2/x^(1/2). Thus, proved.</p>
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<p>We can derive the derivative of 4√x 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: - By First Principle - Using Power Rule - Using Chain Rule We will now demonstrate that the differentiation of 4√x results in 2/x^(1/2) using the above-mentioned methods: By First Principle The derivative of 4√x can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>. To find the derivative of 4√x using the first principle, we will consider f(x) = 4√x. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 4√x, we write f(x + h) = 4√(x + h). Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [4√(x + h) - 4√x] / h We can<a>rationalize</a>the<a>numerator</a>by multiplying by the conjugate: = limₕ→₀ [ (4√(x + h) - 4√x)(4√(x + h) + 4√x) ] / [ h(4√(x + h) + 4√x) ] = limₕ→₀ [ (x + h) - x ] / [ h(4√(x + h) + 4√x) ] = limₕ→₀ h / [ h(4√(x + h) + 4√x) ] = limₕ→₀ 1 / [ 4√(x + h) + 4√x ] = 1 / [ 2√x ] As h approaches 0, 4√(x + h) approaches 4√x, so: f'(x) = 2/x^(1/2). Hence, proved. Using Power Rule To prove the differentiation of 4√x using the<a>power</a>rule, We rewrite the function as 4x^(1/2). By the power rule, d/dx [x^n] = n*x^(n-1), d/dx [4x^(1/2)] = 4 * (1/2) * x^(-1/2) = 2/x^(1/2). Hence, proved. Using Chain Rule We will now prove the derivative of 4√x using the chain rule. The process is as follows: Express the function as a composition: Let u = x^(1/2), then 4√x = 4u. The derivative of u with respect to x is: du/dx = (1/2)x^(-1/2) The derivative of 4u with respect to u is: d/du (4u) = 4 Using the chain rule, d/dx (4u) = d/du (4u) * du/dx = 4 * (1/2)x^(-1/2) = 2/x^(1/2). Thus, proved.</p>
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<h2>Higher-Order Derivatives of 4 Square Root of x</h2>
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<h2>Higher-Order Derivatives of 4 Square Root of x</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.</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.</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 4√x.</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 4√x.</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 4√x, 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 4√x, 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=0, the derivative is undefined because the function involves<a>division by zero</a>. When x=1, the derivative of 4√x = 2/√1 = 2.</p>
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<p>When x=0, the derivative is undefined because the function involves<a>division by zero</a>. When x=1, the derivative of 4√x = 2/√1 = 2.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 4 Square Root of x</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of 4 Square Root of x</h2>
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<p>Students frequently make mistakes when differentiating 4√x. 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 4√x. 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 (4√x · x^2)</p>
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<p>Calculate the derivative of (4√x · 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) = 4√x · x^2.</p>
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<p>Here, we have f(x) = 4√x · x^2.</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 4√x and v = x^2.</p>
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<p>Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 4√x and v = x^2.</p>
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<p>Let’s differentiate each term, u′= d/dx (4√x) = 2/x^(1/2) v′= d/dx (x^2) = 2x</p>
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<p>Let’s differentiate each term, u′= d/dx (4√x) = 2/x^(1/2) v′= d/dx (x^2) = 2x</p>
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<p>Substituting into the given equation, f'(x) = (2/x^(1/2)) · x^2 + (4√x) · 2x</p>
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<p>Substituting into the given equation, f'(x) = (2/x^(1/2)) · x^2 + (4√x) · 2x</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 2x^(3/2) + 8x^(3/2) Therefore, f'(x) = 10x^(3/2).</p>
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<p>Let’s simplify terms to get the final answer, f'(x) = 2x^(3/2) + 8x^(3/2) Therefore, f'(x) = 10x^(3/2).</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>XYZ Construction Company is building a spherical dome. The surface area A of the dome is represented by A = 4π√r, where r is the radius of the dome. If r = 9 meters, find the rate of change of the surface area with respect to the radius.</p>
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<p>XYZ Construction Company is building a spherical dome. The surface area A of the dome is represented by A = 4π√r, where r is the radius of the dome. If r = 9 meters, find the rate of change of the surface area with respect to the radius.</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 A = 4π√r (surface area of the dome)...(1)</p>
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<p>We have A = 4π√r (surface area of the dome)...(1)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative with respect to r: dA/dr = 4π * d/dr (√r) = 4π * (1/2)r^(-1/2) = 2π/r^(1/2)</p>
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<p>Now, we will differentiate the equation (1) Take the derivative with respect to r: dA/dr = 4π * d/dr (√r) = 4π * (1/2)r^(-1/2) = 2π/r^(1/2)</p>
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<p>Given r = 9 (substitute this into the derivative), dA/dr = 2π/√9 dA/dr = 2π/3</p>
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<p>Given r = 9 (substitute this into the derivative), dA/dr = 2π/√9 dA/dr = 2π/3</p>
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<p>Hence, the rate of change of the surface area with respect to the radius when r=9 meters is 2π/3 square meters per meter.</p>
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<p>Hence, the rate of change of the surface area with respect to the radius when r=9 meters is 2π/3 square meters per meter.</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 surface area at r=9 meters as 2π/3, which means that for a small change in the radius, the surface area changes at a rate of 2π/3 square meters per meter.</p>
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<p>We find the rate of change of the surface area at r=9 meters as 2π/3, which means that for a small change in the radius, the surface area changes at a rate of 2π/3 square meters per meter.</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 = 4√x.</p>
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<p>Derive the second derivative of the function y = 4√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 = 2/x^(1/2)...(1)</p>
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<p>The first step is to find the first derivative, dy/dx = 2/x^(1/2)...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2/x^(1/2)]</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [2/x^(1/2)]</p>
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<p>Here, we use the power rule, d²y/dx² = -x^(-3/2)</p>
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<p>Here, we use the power rule, d²y/dx² = -x^(-3/2)</p>
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<p>Therefore, the second derivative of the function y = 4√x is -x^(-3/2).</p>
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<p>Therefore, the second derivative of the function y = 4√x is -x^(-3/2).</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 2/x^(1/2). 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 2/x^(1/2). 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 ((4√x)^2) = 8√x.</p>
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<p>Prove: d/dx ((4√x)^2) = 8√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>Let’s start using the chain rule: Consider y = (4√x)^2 = 16x</p>
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<p>Let’s start using the chain rule: Consider y = (4√x)^2 = 16x</p>
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<p>To differentiate: dy/dx = d/dx (16x) = 16</p>
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<p>To differentiate: dy/dx = d/dx (16x) = 16</p>
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<p>We can also express this using the original form: dy/dx = 2 * (4√x) * d/dx (4√x)</p>
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<p>We can also express this using the original form: dy/dx = 2 * (4√x) * d/dx (4√x)</p>
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<p>Since the derivative of 4√x is 2/x^(1/2), dy/dx = 2 * 4√x * 2/x^(1/2) = 8√x. Hence proved.</p>
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<p>Since the derivative of 4√x is 2/x^(1/2), dy/dx = 2 * 4√x * 2/x^(1/2) = 8√x. 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. We replace 4√x with its derivative and 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. We replace 4√x with its derivative and 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 (4√x/x)</p>
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<p>Solve: d/dx (4√x/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 (4√x/x) = (d/dx (4√x) * x - 4√x * d/dx(x)) / x²</p>
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<p>To differentiate the function, we use the quotient rule: d/dx (4√x/x) = (d/dx (4√x) * x - 4√x * d/dx(x)) / x²</p>
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<p>We will substitute d/dx (4√x) = 2/x^(1/2) and d/dx (x) = 1 = (2/x^(1/2) * x - 4√x) / x² = (2x^(1/2) - 4√x) / x² = (2x^(1/2) - 4x^(1/2)) / x² = -2x^(1/2) / x²</p>
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<p>We will substitute d/dx (4√x) = 2/x^(1/2) and d/dx (x) = 1 = (2/x^(1/2) * x - 4√x) / x² = (2x^(1/2) - 4√x) / x² = (2x^(1/2) - 4x^(1/2)) / x² = -2x^(1/2) / x²</p>
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<p>Therefore, d/dx (4√x/x) = -2/x^(3/2).</p>
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<p>Therefore, d/dx (4√x/x) = -2/x^(3/2).</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 4 Square Root of x</h2>
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<h2>FAQs on the Derivative of 4 Square Root of x</h2>
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<h3>1.Find the derivative of 4√x.</h3>
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<h3>1.Find the derivative of 4√x.</h3>
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<p>Using the power rule for 4√x gives 4x^(1/2): d/dx (4√x) = 2/x^(1/2).</p>
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<p>Using the power rule for 4√x gives 4x^(1/2): d/dx (4√x) = 2/x^(1/2).</p>
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<h3>2.Can we use the derivative of 4√x in real life?</h3>
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<h3>2.Can we use the derivative of 4√x in real life?</h3>
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<p>Yes, we can use the derivative of 4√x in real life to calculate the rate of change of phenomena involving<a>square</a>roots, such as physics and engineering problems.</p>
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<p>Yes, we can use the derivative of 4√x in real life to calculate the rate of change of phenomena involving<a>square</a>roots, such as physics and engineering problems.</p>
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<h3>3.Is it possible to take the derivative of 4√x at x=0?</h3>
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<h3>3.Is it possible to take the derivative of 4√x at x=0?</h3>
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<p>No, x=0 is a point where the function 4√x is undefined due to<a>division</a>by zero, so it is impossible to take the derivative at this point.</p>
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<p>No, x=0 is a point where the function 4√x is undefined due to<a>division</a>by zero, so it is impossible to take the derivative at this point.</p>
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<h3>4.What rule is used to differentiate 4√x/x?</h3>
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<h3>4.What rule is used to differentiate 4√x/x?</h3>
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<p>We use the quotient rule to differentiate 4√x/x: d/dx (4√x/x) = (x * d/dx (4√x) - 4√x * 1) / x².</p>
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<p>We use the quotient rule to differentiate 4√x/x: d/dx (4√x/x) = (x * d/dx (4√x) - 4√x * 1) / x².</p>
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<h3>5.Are the derivatives of 4√x and √(4x) the same?</h3>
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<h3>5.Are the derivatives of 4√x and √(4x) the same?</h3>
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<p>No, they are different. The derivative of 4√x is 2/x^(1/2), while the derivative of √(4x) is 1/√(4x).</p>
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<p>No, they are different. The derivative of 4√x is 2/x^(1/2), while the derivative of √(4x) is 1/√(4x).</p>
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<h2>Important Glossaries for the Derivative of 4 Square Root of x</h2>
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<h2>Important Glossaries for the Derivative of 4 Square Root of x</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function shows 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 shows how the given function changes in response to a slight change in x. </li>
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</ul><ul><li><strong>Square Root Function:</strong>A function that involves the square root of a variable, typically written as √x or x^(1/2). </li>
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</ul><ul><li><strong>Square Root Function:</strong>A function that involves the square root of a variable, typically written as √x or x^(1/2). </li>
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</ul><ul><li><strong>Power Rule:</strong>A basic differentiation rule used to find the derivative of functions of the form x^n. </li>
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</ul><ul><li><strong>Power Rule:</strong>A basic differentiation rule used to find the derivative of functions of the form x^n. </li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating composite functions, allowing us to separate inner and outer functions. </li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating composite functions, allowing us to separate inner and outer functions. </li>
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</ul><ul><li><strong>Undefined Points:</strong>Points where the function does not exist, often due to division by zero or other issues.</li>
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</ul><ul><li><strong>Undefined Points:</strong>Points where the function does not exist, often due to division by zero or other issues.</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>