Let f be a differentiable function defined for all x ∈ R such that f(x3) = x5 for all x ∈ R, x ≠ 0. Then the value of \(\dfrac{df}{dx} (8)\) is:
20/3
Finding the Derivative: Step-by-Step Calculation for \(df/dx\) at \(x=8\)
We are given a differentiable function \(f\) defined for all \(x \in \mathbb{R}\) such that \(f(x^3) = x^5\) for all \(x \in \mathbb{R}, x \neq 0\). We need to find the value of the derivative of \(f\) with respect to \(x\) at \(x=8\), which is denoted as \(\dfrac{df}{dx}(8)\) or \(f'(8)\).
Using the Chain Rule to Differentiate
The given equation is \(f(x^3) = x^5\). To find the derivative \(f'(x)\), we can differentiate both sides of this equation with respect to \(x\). We will use the chain rule on the left side.
Let \(y = x^3\). The left side is \(f(y)\). Differentiating \(f(x^3)\) with respect to \(x\) gives:
\(\dfrac{d}{dx} [f(x^3)] = \dfrac{d}{dy} [f(y)] \cdot \dfrac{dy}{dx}\)
We know that \(\dfrac{d}{dy} [f(y)] = f'(y) = f'(x^3)\), and \(\dfrac{dy}{dx} = \dfrac{d}{dx}(x^3) = 3x^2\). Therefore, the derivative of the left side is:
\(\dfrac{d}{dx} [f(x^3)] = f'(x^3) \cdot 3x^2\)
Now, let's differentiate the right side of the equation \(f(x^3) = x^5\) with respect to \(x\):
\(\dfrac{d}{dx} [x^5] = 5x^{5-1} = 5x^4\)
Equating the derivatives of both sides, we get:
\(f'(x^3) \cdot 3x^2 = 5x^4\)
Solving for \(f'(x^3)\)
We want to find an expression for \(f'(x^3)\). We can isolate \(f'(x^3)\) by dividing both sides by \(3x^2\), assuming \(x \neq 0\):
\(f'(x^3) = \dfrac{5x^4}{3x^2}\)
Simplifying the expression:
\(f'(x^3) = \dfrac{5}{3}x^{4-2} = \dfrac{5}{3}x^2\)
This equation gives us the value of the derivative of \(f\) evaluated at \(x^3\).
Evaluating \(f'(8)\)
We need to find the value of \(f'(8)\). The expression we have is for \(f'(x^3)\). To evaluate \(f'(8)\), we need to find the value of \(x\) such that \(x^3 = 8\).
If \(x^3 = 8\), then \(x\) is the cube root of 8, which is \(x = \sqrt[3]{8} = 2\).
So, to find \(f'(8)\), we need to substitute \(x=2\) into the expression for \(f'(x^3) = \dfrac{5}{3}x^2\).
Substitute \(x=2\):
\(f'(8) = f'(2^3) = \dfrac{5}{3}(2)^2\)
Calculate the value:
\(f'(8) = \dfrac{5}{3}(4)\)
\(f'(8) = \dfrac{20}{3}\)
Thus, the value of \(\dfrac{df}{dx}(8)\) is \(\dfrac{20}{3}\).
Summary of Steps
- Start with the given relation: \(f(x^3) = x^5\).
- Differentiate both sides with respect to \(x\), using the chain rule on the left side: \(\dfrac{d}{dx} [f(x^3)] = \dfrac{d}{dx} [x^5]\).
- This gives \(f'(x^3) \cdot 3x^2 = 5x^4\).
- Solve for \(f'(x^3)\): \(f'(x^3) = \dfrac{5x^4}{3x^2} = \dfrac{5}{3}x^2\).
- To find \(f'(8)\), find the value of \(x\) such that \(x^3 = 8\). This is \(x=2\).
- Substitute \(x=2\) into the expression for \(f'(x^3)\) to find \(f'(8)\): \(f'(8) = \dfrac{5}{3}(2)^2 = \dfrac{20}{3}\).
Revision Table: Key Calculus Concepts
| Concept | Description | Application in this Problem |
|---|---|---|
| Derivative (\(df/dx\) or \(f'(x)\)) | Measures the instantaneous rate of change of a function. | We calculated the derivative of the function \(f(x)\) at a specific point \(x=8\). |
| Chain Rule | Used to differentiate composite functions. If \(h(x) = f(g(x))\), then \(h'(x) = f'(g(x)) \cdot g'(x)\). | Applied to differentiate \(f(x^3)\) with respect to \(x\), where \(f\) is the outer function and \(x^3\) is the inner function. |
| Power Rule of Differentiation | \(\dfrac{d}{dx}(x^n) = nx^{n-1}\). | Used to differentiate \(x^3\) and \(x^5\). |
| Function Evaluation | Finding the value of a function (or its derivative) at a specific input. | We evaluated the derivative \(f'(x)\) at \(x=8\). |
Additional Information: Related Differentiation Techniques
This problem primarily uses the chain rule, but understanding how derivatives work for different function types is crucial in calculus. Here are some related concepts:
- Implicit Differentiation: Used when a function is not explicitly defined in terms of \(x\), but rather through an equation relating \(x\) and \(y\) (where \(y=f(x)\)). You differentiate both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\) and using the chain rule for terms involving \(y\).
- Differentiation of Inverse Functions: If \(y = f(x)\) and \(x = g(y)\) is its inverse, the derivative of the inverse function is given by \(g'(y) = \dfrac{1}{f'(x)}\) or \(\dfrac{dx}{dy} = \dfrac{1}{dy/dx}\), provided \(f'(x) \neq 0\).
- Parametric Differentiation: If \(x\) and \(y\) are both functions of a parameter \(t\), i.e., \(x = x(t)\) and \(y = y(t)\), then \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\), provided \(\dfrac{dx}{dt} \neq 0\).
These techniques extend the power of differentiation to a wider range of functions and relationships.
Important Questions from Differentiability
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