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Chapter 1: Problem 23

Find \(f^{\prime \prime}(x)\). $$f(x)=\sqrt[4]{\left(x^{2}+1\right)^{3}}$$

Short Answer

Expert verified
The second derivative is \(f^{\text{''}}(x) = \frac{3}{2}(x^2 + 1)^{-1/4} - \frac{15x^2}{4}(x^2 + 1)^{-5/4}\).

Step by step solution

01

Rewrite the function in a simpler form

Rewrite the function in exponential form: \[f(x) = \bigg((x^2 + 1)^3\bigg)^{1/4} = (x^2 + 1)^{3/4}\]
02

Find the first derivative \(f^{\text{'}}(x)\)

Use the chain rule to find the first derivative: \[f^{\text{'}}(x) = \frac{d}{dx} \big((x^2 + 1)^{3/4}\big)\] \[= \frac{3}{4}(x^2 + 1)^{-1/4} \frac{d}{dx}(x^2 + 1)\] \[= \frac{3}{4}(x^2 + 1)^{-1/4} (2x)\] \[= \frac{3 \times 2x}{4}(x^2 + 1)^{-1/4}\] \[= \frac{3x}{2}(x^2 + 1)^{-1/4}\]
03

Simplify the first derivative

Simplify the first derivative: \[f^{\text{'}}(x) = \frac{3x}{2}(x^2 + 1)^{-1/4}\]
04

Find the second derivative \(f^{\text{''}}(x)\)

Use the product rule and chain rule to find the second derivative: Let \(u = 3x/2\) and \(v = (x^2 + 1)^{-1/4}\), so \(f^{\text{''}}(x) = u'v + uv'\). First, find \(u'\): \[u' = \frac{d}{dx}\bigg(\frac{3x}{2}\bigg) = \frac{3}{2}\] Second, find \(v'\): \[v' = -\frac{1}{4}(x^2 + 1)^{-5/4} \frac{d}{dx}(x^2 + 1)\] \[= -\frac{5}{4}(x^2 + 1)^{-5/4} (2x)\] \[= -\frac{5x}{2}(x^2 + 1)^{-5/4}\] Then apply the product rule: \[f^{\text{''}}(x) = u'v + uv'\] \[= \frac{3}{2}(x^2 + 1)^{-1/4} + \frac{3x}{2}(-\frac{5x}{2}(x^2 + 1)^{-5/4})\] \[= \frac{3}{2}(x^2 + 1)^{-1/4} - \frac{15x^2}{4}(x^2 + 1)^{-5/4}\]

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Key Concepts

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

calculus
Calculus is the branch of mathematics focusing on the study of change. It consists of two main components, differentiation and integration. In our case, differentiation is concerning the rates of change and slopes of curves.
chain rule
The chain rule is a fundamental differentiation technique used when taking a derivative of a composite function. It states that to find the derivative of a composite function, you multiply the derivative of the outer function by the derivative of the inner function. In simpler terms, you 'chain' these derivatives together, which is crucial for dealing with nested functions.
product rule
The product rule is essential for differentiating products of two or more functions. It states that if you have two functions multiplied together, the derivative is found by taking the derivative of the first function and multiplying it by the second function, plus the first function times the derivative of the second function. This allows you to handle more complex expressions efficiently.
differentiation
Differentiation is the process of finding a derivative, which measures how a function changes as its input changes. It is used to determine rates of change, tangents to curves, and slopes of functions. The key techniques include the chain rule, product rule, and the power rule, among others. Mastering these rules is essential for solving complex calculus problems.

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Most popular questions from this chapter

Find \(d y / d x .\) Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$y=\frac{x^{5}-3 x^{4}+2 x+4}{x^{2}}$$

A proof of the Product Rule appears below. Provide a justification for each step. $$\text { a) } \frac{d}{d x}[f(x) \cdot g(x)]=\lim _{h \rightarrow 0} \frac{f(x+h) g(x+h)-f(x) g(x)}{h}$$ $$\text { b) } \quad=\lim _{h \rightarrow 0} \frac{f(x+h) g(x+h)-f(x+h) g(x)+f(x+h) g(x)-f(x) g(x)}{h}$$ $$\text { c) } \quad=\lim _{h \rightarrow 0} \frac{f(x+h) g(x+h)-f(x+h) g(x)}{h}+\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)}{h}$$ $$\begin{array}{ll} \text { d) } & =\lim _{h \rightarrow 0}\left[f(x+h) \cdot \frac{g(x+h)-g(x)}{h}\right]+\lim _{h \rightarrow 0}\left[g(x) \cdot \frac{f(x+h)-f(x)}{h}\right] \end{array}$$ $$\text { e) } \quad=f(x) \cdot \lim _{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}+g(x) \cdot \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ $$\mathbf{f}) \quad=f(x) \cdot g^{\prime}(x)+g(x) \cdot f^{\prime}(x)$$ $$\text { g) } \quad=f(x) \cdot\left[\frac{d}{d x} g(x)\right]+g(x) \cdot\left[\frac{d}{d x} f(x)\right]$$

First, use the Chain Rule to find the answer. Next, check your answer by finding \(f(g(x))\) taking the derivative, and substituting. \(f(u)=u^{3}, g(x)=u=2 x^{4}+1\) Find \((f \circ g)^{\prime}(-1) .\)

Use the Chain Rule to differentiate each function. You may need to apply the rule more than once. $$f(x)=\left(2 x^{5}+(4 x-5)^{2}\right)^{6}$$

Use the Chain Rule to differentiate each function. You may need to apply the rule more than once. $$f(x)=\left(-x^{5}+4 x+\sqrt{2 x+1}\right)^{3}$$
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