/*! 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 16 Find the derivative. $$ \fra... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the derivative. $$ \frac{1}{\sqrt{x+1}} $$

Short Answer

Expert verified
The derivative is \( -\frac{1}{2}(x+1)^{-3/2} \).

Step by step solution

01

Rewrite the Expression

The given expression is \( \frac{1}{\sqrt{x+1}} \). Rewrite it using exponent notation: \( (x+1)^{-1/2} \). This makes it easier to apply differentiation rules.
02

Apply the Power Rule

To differentiate an expression of the form \( x^n \), use the power rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \). Here, let \( u = x + 1 \) and use \( n = -\frac{1}{2} \). Start by differentiating \( u^{-1/2} \) as: \( \frac{d}{du}(u^{-1/2}) = -\frac{1}{2}u^{-3/2} \).
03

Apply the Chain Rule

Use the chain rule to account for the inner function \( u = x+1 \). The chain rule states \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \). Here, \( \frac{du}{dx} = 1 \). So the derivative with respect to \( x \) is: \( -\frac{1}{2}(x+1)^{-3/2} \times 1 = -\frac{1}{2}(x+1)^{-3/2} \).

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Key Concepts

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

Understanding the Chain Rule
The chain rule is a powerful and essential tool in differentiation. It helps us to differentiate composite functions effectively by handling the inside and outside functions separately. When you have a function nested inside another function, such as \( f(g(x)) \), you need the chain rule to find the derivative of the whole expression.

The chain rule formula is:
  • \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
Here, \( u \) is an intermediate variable representing the inside function \( g(x) \). To apply the chain rule, follow these steps:
  • Identify the inner function (\( u \)) and the outer function (\( f \)).
  • Ddifferentiate the outer function with respect to \( u \), and the inner function with respect to \( x \).
  • Multiply these derivatives together to obtain the derivative of the composite function.
In our problem, we set \( u = x + 1 \), making it easier to rewrite and differentiate the expression. The chain rule allows us to properly handle the additional differentiation step due to the inner function \( u \). This ensures we account for the change in \( x \) across both functions involved.
Mastering the Power Rule
The power rule is a fundamental differentiation technique used to derive powers of variables efficiently. It's broadly used when you have a function of the form \( x^n \).The power rule is straightforward and is expressed as follows:
  • \( \frac{d}{dx}(x^n) = nx^{n-1} \)
This means that you multiply by the current power and subtract one from the exponent to find the derivative.

In our specific exercise, we've rewritten the original function as \( (x+1)^{-1/2} \), making it suitable for applying the power rule. Here, the base \( (x+1) \) acts similarly to our \( x \), and the exponent \( n \) is \(-1/2 \). By applying the power rule, we bring down the exponent and modify it:
  • The derivative with respect to \( u \) becomes \( -\frac{1}{2}u^{-3/2} \).
This step reduces the power by one, allowing us to reach the point where the chain rule can be seamlessly applied to complete the differentiation process.
Applying Differentiation Techniques
Differentiation is a core concept of calculus, providing a way to compute the rate at which a function changes. In calculus, powerful techniques like the chain rule and power rule allow us to differentiate functions efficiently and accurately.

The combination of differentiation techniques, as shown in our solution, enables us to handle more complex functions composed of simpler elements. By initiating with the power rule to intervene on the observable power, the function becomes manageable. Then, the chain rule precisely tackles the nested element inside the function, ensuring all variational changes are considered.

This comprehensive toolkit—a combination of rules—empowers you to differentiate a broad spectrum of expressions. Starting from rewriting complicated functions in differentiation-friendly forms, these strategies streamline the process and enhance your calculus proficiency. This is why understanding and applying each rule on its own, and in combination, is vital for mastering calculus.

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Most popular questions from this chapter

74\. Lamprey Growth Griffiths and colleagues \(^{8}\) studied the larval sea lamprey in the Great Lakes, where they represent a serious threat to the region's fisheries. They created a mathematical model given by the equation \(G(T)=-0.991+0.273 T-0.0179 T^{2},\) where \(T\) is the mean annual water temperature with \(6.2 \leq T \leq 9.8\) and \(G\) is the growth in millimeters per day. Find \(G^{\prime}(T) .\) Find \(G^{\prime}(8) .\) Give units.

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