Problem 120
Question
If \(g\) is the inverse of a function \(f\) and \(f^{\prime}(x)=\frac{1}{1+x^{5}}\), then \(g^{\prime}(x)\) is equal to (A) \(1+x^{5}\) (B) \(5 x^{4}\) (C) \(\frac{1}{1+\\{g(x)\\}^{5}}\) (D) \(1+\\{g(x)\\}^{5}\)
Step-by-Step Solution
Verified Answer
The answer is (D) \( 1 + g(x)^{5} \).
1Step 1: Understand the Relationship between Inverse Functions and Derivatives
When g is the inverse of f, then for any x, if y = g(x) then x = f(y). We know the relationship between the derivatives of inverse functions: if g is the inverse of f, then \( g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))} \).
2Step 2: Identify Given Derivative
We are given that \( f^{\prime}(x) = \frac{1}{1+x^{5}} \). Our task is to find \( g^{\prime}(x) \) using the relationship identified earlier.
3Step 3: Apply the Derivative of Inverse Relationship
According to the formula, \( g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))} \). Substitute the expression for \( f^{\prime}(x) \), so \( g^{\prime}(x) = \frac{1}{\frac{1}{1+g(x)^{5}}} \).
4Step 4: Simplify the Expression
The equation simplifies to \( g^{\prime}(x) = 1 + g(x)^{5} \) since \( \frac{1}{\frac{1}{1+g(x)^{5}}} = 1 + g(x)^{5} \).
5Step 5: Choose the Correct Option
Refer back to the options given. This expression, \( g^{\prime}(x) = 1 + g(x)^{5} \), corresponds to option D.
Key Concepts
Derivatives of Inverse FunctionsFunction DerivativesInverse Function Derivative Formula
Derivatives of Inverse Functions
Understanding the derivatives of inverse functions might seem tricky at first, but it's quite fascinating! Here’s the scoop: if you have a function, say \( f \), with an inverse function \( g \), the derivatives of these functions are connected in a unique way. This relationship helps us understand how changes in one function reflect as changes in its inverse.
Here's the magic formula for their derivatives: if \( g \) is the inverse of \( f \), then \( g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))} \).
Let's visualize this as a see-saw. If one side changes, the other side adjusts in a specific reciprocal manner. This derivative relationship is essential in calculus as it allows us to easily find the derivative of an inverse function given the derivative of the original function.
Here's the magic formula for their derivatives: if \( g \) is the inverse of \( f \), then \( g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))} \).
Let's visualize this as a see-saw. If one side changes, the other side adjusts in a specific reciprocal manner. This derivative relationship is essential in calculus as it allows us to easily find the derivative of an inverse function given the derivative of the original function.
Function Derivatives
Function derivatives are like the speedometers of mathematics—they tell us how fast or slow a function is changing at any given point in its domain. In a more technical sense, the derivative of a function at any point gives us the slope of the tangent line to the graph of the function at that point.
For instance, for a function \( f(x) \), its derivative \( f^{\prime}(x) \) is calculated using the limit process:
For instance, for a function \( f(x) \), its derivative \( f^{\prime}(x) \) is calculated using the limit process:
- First, consider a small change in \( x\), then observe the change in \( f(x) \).
- Use the limit \( \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \) to find the derivative.
Inverse Function Derivative Formula
The derivative formula for inverse functions is a powerful tool. When you need to find the derivative of an inverse function, this formula makes the task much simpler. Let's break it down: Suppose \( g \) is the inverse of a function \( f \) and we need \( g^{\prime}(x) \). You can use the derivative of \( f \) to find it with this formula: \( g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))} \).
In our specific scenario, we're given \( f^{\prime}(x) = \frac{1}{1+x^{5}} \). Plugging this into the inverse function derivative formula leads us to:
In our specific scenario, we're given \( f^{\prime}(x) = \frac{1}{1+x^{5}} \). Plugging this into the inverse function derivative formula leads us to:
- Substitute \( g(x)\), resulting in \( g^{\prime}(x) = \frac{1}{\frac{1}{1+g(x)^{5}}} \).
- Simplify this expression to get \( g^{\prime}(x) = 1 + g(x)^{5} \).
Other exercises in this chapter
Problem 118
\(\frac{d^{2} x}{d y^{2}}\) is equal to (A) \(-\left(\frac{d^{2} y}{d x^{2}}\right)^{-1}\left(\frac{d y}{d x}\right)^{-3}\) (B) \(\left(\frac{d^{2} y}{d x^{2}}\
View solution Problem 119
If \(y=\sec \left(\tan ^{-1} x\right)\), then \(\frac{d y}{d x}\) at \(x=1\) is equal to (A) \(\frac{1}{2}\) (B) 1 (C) \(\sqrt{2}\) (D) \(\frac{1}{\sqrt{2}}\)
View solution Problem 117
Let \(f:(-1,1) \rightarrow R\) be a differentiable function such that \(f(0)=-1\) and \(f^{\prime}(0)=1\). Let \(g(x)=[f(2 f(x)+2)]^{2}\). Then \(g^{\prime}(0)=
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