/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Consider the following functions... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following functions. In each case, without finding the inverse, evaluate the derivative of the inverse at the given point. $$f(x)=\ln (5 x+e) ;(1,0)$$

Short Answer

Expert verified
Question: Find the derivative of the inverse function, \(f^{-1}(x)\), at the point (1,0) without finding the inverse function, given that \(f(x)=\ln(5x+e)\). Answer: The derivative of the inverse function at the point (1,0) is \((f^{-1})'(1)=\frac{e}{5}\).

Step by step solution

01

Find the derivative of f(x)

To find the derivative of the given function \(f(x)=\ln(5x+e)\), we can use the chain rule. The chain rule states that if we have a function g(h(x)), then its derivative is given by: $$\frac{d}{dx}g(h(x))=g'(h(x))\cdot h'(x)$$ In our case, \(g(u)=\ln(u)\) with the derivative \(g'(u)=\frac{1}{u}\) and \(h(x)=5x+e\) with the derivative \(h'(x)=5\). Applying the chain rule, we have: $$f'(x)=g'(h(x))\cdot h'(x)=\frac{1}{5x+e}\cdot 5$$
02

Find f'(f^{-1}(1))

Using the given point \((1,0)\), we know that \(f^{-1}(1)=0\). So we need to evaluate \(f'(0)\): $$f'(0)=\frac{1}{5\cdot0+e}\cdot 5=\frac{1}{e}\cdot 5$$
03

Find the derivative of the inverse function

Using the formula for the derivative of the inverse function, we have: $$(f^{-1})'(1)=\frac{1}{f'(f^{-1}(1))}=\frac{1}{\frac{5}{e}}=\frac{e}{5}$$ So, the derivative of the inverse function at the point \((1,0)\) is \((f^{-1})'(1)=\frac{e}{5}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is a fundamental concept in calculus that helps us differentiate composite functions.
Think of a composite function as a function within another function. When we want to find the derivative of such a function, we apply the chain rule.
Here's how it works:
  • Let's say you have a composite function, denoted as \(g(h(x))\). This means \(g(x)\) is applied to \(h(x)\).
  • The chain rule tells us that the derivative \( \frac{d}{dx}g(h(x)) \) is the product of the derivative of \(g\) with respect to \(h\) times the derivative of \(h\) with respect to \(x\).
To illustrate this with our original function example, we have \(f(x) = \ln(5x + e)\). Here, \(\ln(u)\) is our outer function \(g\), where \(u = 5x + e\) is the inner function \(h(x)\).
Using the chain rule, we differentiate each part:
  • First, differentiate the outer function: \(g'(u) = \frac{1}{u}\).
  • Then, differentiate the inner function: \(h'(x) = 5\).
Multiply these derivatives together to find the derivative of the whole function, \(f'(x) = \frac{1}{5x + e} \cdot 5\). This is the essence of the chain rule, allowing us to break down complex differentiation into simpler parts.
Natural Logarithm Function
The natural logarithm function \(\ln(x)\) is one of the most important functions in mathematics, especially in calculus.
It acts as the inverse of the exponential function with base \(e\), where \(e\) is approximately 2.71828.
This means that \(e^{\ln(x)} = x\) and \(\ln(e^x) = x\).
  • The natural logarithm function is unique for its natural base \(e\), making it particularly useful in calculus for its simple derivative properties.
  • Another important property of \(\ln(x)\) is that it is undefined for non-positive values, which means it only takes positive real numbers as inputs.
When differentiating \(\ln(x)\), the result is a clean and simple expression: \(\frac{d}{dx}\ln(x) = \frac{1}{x}\).
This derivative is especially simple compared to other logarithmic derivatives, which makes the natural logarithm a favorite choice.
In our problem, the natural logarithm function appears, and it guides us when applying the chain rule by simplifying the derivation of our composite function \(\ln(5x + e)\). Natural logarithms reveal their strength in calculus by transforming multiplication into addition (property of logarithms), which greatly simplifies many equations as seen in problems involving compounded growth or decay.
Derivative Using Chain Rule
After understanding the chain rule and the properties of the natural logarithm, we can effectively find the derivative using the chain rule.
In our example, with the function \(f(x) = \ln(5x + e)\), we aim to find the derivative and further the derivative of its inverse.
The chain rule is the perfect tool here:
  • We first find \(f'(x)\) by differentiating \(\ln(5x + e)\) using the steps covered with the chain rule, resulting in \(\frac{5}{5x + e}\).
  • This derivative helps to find critical behavior in functions, such as points where the slope of the tangent is zero, indicating minima or maxima.
Once \(f'(x)\) is known, we use it to find the derivative of the inverse function, \((f^{-1})'(x)\).
The inverse function derivative is given by \((f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\).
In this problem, such evaluation at the point \((1, 0)\) gives us \((f^{-1})'(1) = \frac{e}{5}\). This demonstrates the handy use of derivatives in understanding how inverse functions change with respect to inputs.Overall, derivatives, especially using techniques like the chain rule, allow us to unravel the behavior of complex functions and their inverses, providing insights into mathematical models and real-world phenomena.

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Most popular questions from this chapter

Let \(f(x)=x e^{2 x}\) a. Find the values of \(x\) for which the slope of the curve \(y=f(x)\) is 0 b. Explain the meaning of your answer to part (a) in terms of the graph of \(f\)

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Determine whether the following statements are true and give an explanation or counterexample. a. For any equation containing the variables \(x\) and \(y,\) the derivative \(d y / d x\) can be found by first using algebra to rewrite the equation in the form \(y=f(x).\) b. For the equation of a circle of radius \(r, x^{2}+y^{2}=r^{2},\) we have \(\frac{d y}{d x}=-\frac{x}{y},\) for \(y \neq 0\) and any real number \(r>0.\) c. If \(x=1\), then by implicit differentiation, \(1=0.\) d. If \(x y=1,\) then \(y^{\prime}=1 / x.\)

a. Determine the points where the curve \(x+y^{3}-y=1\) has a vertical tangent line (see Exercise 60 ). b. Does the curve have any horizontal tangent lines? Explain.

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