Problem 18
Question
Let \(f\) be a differentiable function satisfying \([f(x)]^{n}=f(n x)\) for all \(x \in R\). Then, \(f^{\prime}(x) f(n x)\) (a) \(f(x)\) (b) 0 (c) \(f(x) f^{\prime}(n x)\) (d) None of these
Step-by-Step Solution
Verified Answer
(c) \(f(x) f^{\prime}(n x)\)
1Step 1: Differentiate both sides of the equation
Given the equation \([f(x)]^{n}=f(n x)\), we start by differentiating both sides with respect to \(x\). This involves using the chain rule for differentiation. The left side becomes \(n[f(x)]^{n-1}f'(x)\) by applying the chain rule, and the right side yields \(n f'(n x)\) because we differentiate \(f(n x)\) with respect to \(x\). Thus, the equation becomes: \[ n[f(x)]^{n-1}f'(x) = n f'(n x). \]
2Step 2: Simplify the differentiated equation
To simplify the equation \( n[f(x)]^{n-1}f'(x) = n f'(n x) \), we can divide both sides by \(n\), assuming \(n eq 0\). This simplifies the equation to: \[ [f(x)]^{n-1}f'(x) = f'(n x). \]
3Step 3: Multiply by \(f(n x)\)
Now, multiply both sides of the simplified equation \([f(x)]^{n-1}f'(x) = f'(n x)\) by \(f(n x)\). This will give: \[ f(n x)[f(x)]^{n-1}f'(x) = f'(n x)f(n x). \]
4Step 4: Substitution and Simplification
Recognize that \([f(x)]^{n} = f(n x)\) from the original equation. This implies \(f(n x)[f(x)]^{n-1} = [f(x)]^{n}\). Substitute this back into our multiplied equation: \[ [f(x)]^{n}f'(x) = f'(n x)f(n x). \] The expression on the left-hand side simplifies to \(f(n x)f'(x)\).
5Step 5: Verify the solution
Our goal is to express \(f^{\prime}(x) f(n x)\) using the previously derived equalities. The equality \(f(n x)f'(x) = f'(n x)f(n x)\) shows that the left side is the same as the right side, confirming the identities required by the problem. Hence, \(f'(x)f(n x) = f(x)f'(n x)\).
Key Concepts
Differentiable FunctionChain RuleDifferentiationSolving Equations
Differentiable Function
A differentiable function is a critical concept in calculus, especially in differential calculus. It is a function that has a derivative at each point in its domain. This means you can find a tangent line at any point on its curve.
A differentiable function is smooth and continuous. There won’t be any sharp turns or cusps in the graph of the function. If a function is differentiable at a point, it's also continuous at that point, but the reverse is not always true.
Understanding differentiability helps in solving problems where you need to find the rate of change. It's also essential in optimization problems, where you're trying to find the maximum or minimum values of a function.
A differentiable function is smooth and continuous. There won’t be any sharp turns or cusps in the graph of the function. If a function is differentiable at a point, it's also continuous at that point, but the reverse is not always true.
Understanding differentiability helps in solving problems where you need to find the rate of change. It's also essential in optimization problems, where you're trying to find the maximum or minimum values of a function.
Chain Rule
The chain rule is a powerful and essential tool in differentiation. It allows us to differentiate composite functions. For a composite function of the form \(f(g(x))\), the chain rule states that the derivative \(f'(x)\) is \(f'(g(x)) \cdot g'(x)\).
This means you first take the derivative of the outer function (\(f\)), keeping the inner function unchanged, and then multiply by the derivative of the inner function (\(g\)).
In the context of the original exercise, using the chain rule was vital to successfully differentiate the given equation \([f(x)]^{n}=f(n x)\) with respect to \(x\). The chain rule allowed us to work with expressions involving nested functions, making the differentiation process systematic and straightforward.
This means you first take the derivative of the outer function (\(f\)), keeping the inner function unchanged, and then multiply by the derivative of the inner function (\(g\)).
In the context of the original exercise, using the chain rule was vital to successfully differentiate the given equation \([f(x)]^{n}=f(n x)\) with respect to \(x\). The chain rule allowed us to work with expressions involving nested functions, making the differentiation process systematic and straightforward.
Differentiation
Differentiation is the process of finding a derivative, which measures how a function changes as its input changes. It is one of the core operations in calculus. The derivative of a function at a point gives the slope of the tangent to the curve at that point.
Differentiation allows us to compute these rates of change and is fundamental in understanding the behavior of functions. It aids in solving various problems, ranging from physics to economics.
In the exercise, differentiation was applied to both sides of the given equation. This led to expressions that could be further simplified, allowing us to explore relationships between functions and their derivatives.
Differentiation allows us to compute these rates of change and is fundamental in understanding the behavior of functions. It aids in solving various problems, ranging from physics to economics.
In the exercise, differentiation was applied to both sides of the given equation. This led to expressions that could be further simplified, allowing us to explore relationships between functions and their derivatives.
Solving Equations
Solving equations often involves finding conditions under which equations hold true. In calculus, this frequently includes equations involving derivatives. The key is to manage these equations systematically, often by simplifying and substituting expressions.
In the step-by-step solution, solving played a role in expressing the problem's requirements in terms of derivatives. Simplification allowed the equation to reveal insights about the relationships between the parts of a function after differentiation.
In the step-by-step solution, solving played a role in expressing the problem's requirements in terms of derivatives. Simplification allowed the equation to reveal insights about the relationships between the parts of a function after differentiation.
- Start by differentiating or transforming terms to simplify the equation.
- Use algebraic manipulation to isolate and solve for desired expressions.
Other exercises in this chapter
Problem 16
The functions \(u=e^{x} \sin x, v=e^{x} \cos x\) satisfy the equation (a) \(v \frac{d u}{d x}-u \frac{d v}{d x}=u^{2}+v^{2}\) (b) \(\frac{d^{2} u}{d x^{2}}=2 v\
View solution Problem 17
If \(f(x)=\log _{x}\\{\ln (x)\\}\), then \(f^{\prime}(x)\) at \(x=e\), is (a) \(e\) (b) \(-e\) (c) \(e^{2}\) (d) \(e^{-1}\)
View solution Problem 19
If \(y=f(x)\) is an odd differentiable function defined on \((-\infty, \infty)\) such that \(f^{\prime}(3)=-2\), then \(f^{\prime}(-3)\) equals (a) 4 (b) 2 \(\b
View solution Problem 21
If \(y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots\left(1+x^{2^{n}}\right)\), then \(\frac{d y}{d x}\) at \(x=0\) is (a) 1 (b) \(-1\) (c) 0 (d) None of
View solution