/*! 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 15 Express \(d^{2} y / d x^{2}\) in... [FREE SOLUTION] | 91Ó°ÊÓ

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Express \(d^{2} y / d x^{2}\) in terms of \(x\) and \(y\). \(4 \tan y=x^{3}\).

Short Answer

Expert verified
\(d^{2} y / d x^{2}=\frac{-8x^{5}+24x^{7}}{(4+4x^{6})^{2}}\)

Step by step solution

01

Express y in terms of x

From \(4 \tan y=x^{3}\), one can express \(y\) in terms of \(x\) as follows: \(y=\tan^{-1}\left(\frac{x^{3}}{4}\right)\)
02

Find \(dy/dx\)

The derivative of \(y\) with respect to \(x\) can be found using the chain rule: \[\frac{dy}{dx}=\frac{1}{1+\left(\frac{x^{3}}{4}\right)^{2}} \times \frac{d}{dx}\left(\frac{x^{3}}{4}\right)=\frac{1}{1+\frac{x^{6}}{16}}\times\frac{3x^{2}}{4}\] Simplifying, this gives \[\frac{dy}{dx}=\frac{3x^{2}}{4+4x^{6}}\]
03

Find \(d^{2}y/dx^{2}\)

The second derivative is found by differentiating \(\frac{dy}{dx}\) with respect to \(x\). Using standard differentiation rules for fractions (quotient rule) we get: \[\frac{d^{2}y}{dx^{2}}=\frac{(8x^{5}+24x^{7})*\frac{3x^{2}}{4+4x^{6}}-(6x-24x^{6})*\frac{4}{4+4x^{6}}}{(4+4x^{6})^{2}} =\frac{24x^{7}+8x^{5}-24x^{6}x^{2}}{(4+4x^{6})^{2}}.\] Then simplifying this gives us: \[\frac{d^{2}y}{dx^{2}}=\frac{-8x^{5}+24x^{7}}{(4+4x^{6})^{2}}.\]

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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 tool in calculus used to differentiate composite functions. It helps us find how one function changes when there’s a "function inside another function." Consider the composite function as two linked pieces: an outer function and an inner function.

For instance, if you have a function \( y = \tan^{-1}\left(\frac{x^3}{4}\right) \), the inner function is \( u = \frac{x^3}{4} \) and the outer function is \( \tan^{-1}(u) \).

To differentiate, you follow these steps:
  • Find the derivative of the outer function with respect to the inner function.
  • Multiply it by the derivative of the inner function with respect to the variable \( x \).
This is mathematically represented as \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
This simple formula is the essence of the Chain Rule and makes handling complex functions manageable.
Quotient Rule
The Quotient Rule is particularly useful when you're dealing with derivatives of fractions. It helps to find the derivative of a function that is the ratio of two differentiable functions. Given two functions \( u(x) \) and \( v(x) \) where the function is \( \frac{u(x)}{v(x)} \), the quotient rule states:
\[ \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]

In our exercise, you apply the Quotient Rule to differentiate \( \frac{dy}{dx} = \frac{3x^2}{4+4x^6} \), which involves treating the numerator and denominator as separate functions.

Let's break it down:
  • Differentiate the numerator: \( \frac{d}{dx}(3x^2) \).
  • Differentiate the denominator: \( \frac{d}{dx}(4+4x^6) \).
  • Use the Quotient Rule to combine these derivatives.
The result gives the second derivative, which helps us understand the curvature and concavity of the original curve.
Second Derivative
The second derivative provides deeper insights into the behavior of functions. It is essentially the derivative of the first derivative, meaning it tells us how the rate of change itself is changing.

When calculating \( \frac{d^2y}{dx^2} \) for our function, you start by taking the derivative of \( \frac{dy}{dx} \) again.

For practical purposes, the second derivative can tell you about:
  • Concavity: If \( \frac{d^2y}{dx^2} > 0 \), the graph is concave up.
  • Convexity: If \( \frac{d^2y}{dx^2} < 0 \), the graph is concave down.
This is key in various applications like graphing curves or analyzing motion as it provides information beyond just simple direction change.
Implicit Differentiation
Implicit Differentiation is a technique used when a function is not easily solved for one variable in terms of another. It means you differentiate both sides of an equation simultaneously, keeping in mind that one of the variables is implicit.

In our exercise, the equation \( 4 \tan y = x^3 \) can't be easily expressed as \( y = f(x) \) initially.

So, how does implicit differentiation work? Here’s the process:
  • Differentially both sides of the equation with respect to \( x \).
  • Apply differentiation rules, such as the Chain Rule, wherever necessary because \( y \) is a function of \( x \).
  • Solve for \( \frac{dy}{dx} \).
This technique is powerful for equations where the dependent variable isn't explicitly isolated, allowing you to still find derivatives as needed.

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Most popular questions from this chapter

Show that the sum of the \(x\) - and \(y\) -intercepts of any line tangent to the graph of \(x^{1 / 2}+y^{1 / 2}=c^{-2}\) is constant and equal to \(c\).

Find conditions on \(a, b, c, d\) which guarantee that the graph of the cubic \(p(x)=a x^{3}+b x^{2}+c x+d\) has: (a) exactly two horizontal tangents. (b) exactly one horizontal tangent. (c) no horizontal tangents.

Set \(f(x)=3 \sqrt[3]{x} .\) Use a CAS to (a) Find \(d(h)=\frac{f(h)-f(0)}{h}\). (b) Find \(\lim _{h \rightarrow 0^{-}} d(h)\) and \(\lim _{h \rightarrow 0^{+}} d(h)\) (c) Is there a tangent line at (0.0)\(?\) Explain. (d) Use a graphing utility to draw the graph of \(f\) on [-2,2].

Set \(f(x)=\frac{1}{1+x^{2}}\) (a) Use a CAS to find \(f^{\prime}(1)\). Then find an equation for the line \(l\) tangent to the graph of \(f\) at the point \((1, f(1))\) (b) Use a graphing utility to display \(l\) and the graph of \(f\) in one figure. (c) Note that \(l\) is a p.sod approximation to the graph of \(f\) for \(x\) close to \(1 .\) Determine the interval on which the vertical separation between \(l\) and the graph of \(f\) is of absolute value less than 0.01

Find an equation for the line \(l\) tangent to the graph of \(f\) at the point with \(x\) -coordinate \(c\). Use a graphing utility to display / and the graph of \(f\) in one figure. Note that \(l\) is a good approximation to the graph of \(f\) for \(x\) close to \(c\). Determine the interval on which the vertical separation between \(l\) and the graph of \(f\) is of absolute value less than 0.01. $$f(x)=\tan x ; c=\pi / 4$$

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