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Use the formula \( f^{\prime}(x)=\lim _{z \rightarrow x} \frac{f(z)-f(x)}{z-x} \) to find the derivative of the following function \( \quad f(x)=\frac{4}{x+4} \)

Ask by Park Williams. in the United States
Feb 03,2025

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Answer

The derivative of \( f(x) = \frac{4}{x + 4} \) is \( f'(x) = -\frac{4}{(x + 4)^2} \).

Solution

To find the derivative of the function \( f(x) = \frac{4}{x + 4} \) using the limit definition of the derivative, we'll follow these steps: \[ f'(x) = \lim_{z \to x} \frac{f(z) - f(x)}{z - x} \] Let's proceed step-by-step. ### Step 1: Substitute \( f(z) \) and \( f(x) \) into the Limit Given: \[ f(z) = \frac{4}{z + 4} \quad \text{and} \quad f(x) = \frac{4}{x + 4} \] Substitute these into the limit: \[ f'(x) = \lim_{z \to x} \frac{\frac{4}{z + 4} - \frac{4}{x + 4}}{z - x} \] ### Step 2: Combine the Fractions in the Numerator To combine the two fractions in the numerator, find a common denominator: \[ \frac{4}{z + 4} - \frac{4}{x + 4} = \frac{4(x + 4) - 4(z + 4)}{(z + 4)(x + 4)} \] Simplify the numerator: \[ 4(x + 4) - 4(z + 4) = 4x + 16 - 4z - 16 = 4x - 4z = 4(x - z) \] So now the expression becomes: \[ f'(x) = \lim_{z \to x} \frac{4(x - z)}{(z + 4)(x + 4)(z - x)} \] ### Step 3: Simplify the Expression Notice that \( x - z = -(z - x) \), so: \[ 4(x - z) = -4(z - x) \] Substitute this back into the limit: \[ f'(x) = \lim_{z \to x} \frac{-4(z - x)}{(z + 4)(x + 4)(z - x)} = \lim_{z \to x} \frac{-4}{(z + 4)(x + 4)} \] ### Step 4: Evaluate the Limit As \( z \to x \), \( z + 4 \to x + 4 \). Therefore: \[ f'(x) = \frac{-4}{(x + 4)(x + 4)} = \frac{-4}{(x + 4)^2} \] ### Final Answer \[ f'(x) = -\frac{4}{(x + 4)^2} \]

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To find the derivative of the function \( f(x) = \frac{4}{x+4} \) using the formula for the derivative, we start by substituting into the formula: \[ f^{\prime}(x) = \lim_{z \rightarrow x} \frac{f(z) - f(x)}{z - x} \] First, compute \( f(z) \): \[ f(z) = \frac{4}{z + 4} \] Now, compute \( f(z) - f(x) \): \[ f(z) - f(x) = \frac{4}{z + 4} - \frac{4}{x + 4} \] To combine these fractions, find a common denominator: \[ = \frac{4(x + 4) - 4(z + 4)}{(z + 4)(x + 4)} = \frac{4x + 16 - 4z - 16}{(z + 4)(x + 4)} = \frac{4x - 4z}{(z + 4)(x + 4)} = \frac{4(x - z)}{(z + 4)(x + 4)} \] Now substitute back into the limit: \[ f^{\prime}(x) = \lim_{z \rightarrow x} \frac{\frac{4(x - z)}{(z + 4)(x + 4)}}{z - x} \] This simplifies to: \[ = \lim_{z \rightarrow x} \frac{4(x - z)}{(z + 4)(x + 4)(z - x)} = \lim_{z \rightarrow x} \frac{-4}{(z + 4)(x + 4)} \] Because \( z - x = -(x - z) \). Now, as \( z \) approaches \( x \): \[ f^{\prime}(x) = -\frac{4}{(x + 4)(x + 4)} = -\frac{4}{(x + 4)^2} \] So the derivative is: \[ f^{\prime}(x) = -\frac{4}{(x + 4)^2} \]

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