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Chapter 52: Chapter 52

Problem 970

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Find the derivative of: \(f(x)=\left(x^{2}+1\right) /(1-3 x)\).

Short Answer

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The short answer for the derivative of \(f(x) = \frac{x^2 + 1}{1 - 3x}\) is \(f'(x) = \frac{-3x^2 + 2x + 3}{(1 - 3x)^2}\).
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Identify the numerator and denominator functions

We have a function in the form of \(\frac{u(x)}{v(x)}\), where \(u(x) = x^2 + 1\) and \(v(x) = 1 - 3x\).

Step 2: Calculate the derivatives of the numerator and denominator functions

Now we will find the derivatives of the two functions - \(u'(x)\) and \(v'(x)\). For \(u(x) = x^2 + 1\), use the Power Rule: \[u'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(1) = 2x + 0 = 2x\] For \(v(x) = 1 - 3x\), again, use the Power Rule: \[v'(x) = \frac{d}{dx}(1) - \frac{d}{dx}(3x) = 0 - 3 = -3\]

Step 3: Apply the Quotient Rule

Now that we have found \(u'(x)\) and \(v'(x)\), we can apply the Quotient Rule to find the derivative of \(f(x)\). The Quotient Rule states: \[\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\] Plug in the expressions we found in Steps 1 and 2: \[\frac{d}{dx}\left(\frac{x^2 + 1}{1 - 3x}\right) = \frac{(2x)(1 - 3x) - (x^2 + 1)(-3)}{(1 - 3x)^2}\]

Step 4: Simplify the expression

Now, simplify the expression obtained in Step 3: \[\frac{d}{dx}\left(\frac{x^2 + 1}{1 - 3x}\right) = \frac{2x(1 - 3x) + 3(x^2 + 1)}{(1 - 3x)^2}\] Expand the numerator and combine like terms: \[\frac{d}{dx}\left(\frac{x^2 + 1}{1 - 3x}\right) = \frac{2x - 6x^2 + 3x^2 + 3}{(1 - 3x)^2}\] \[\frac{d}{dx}\left(\frac{x^2 + 1}{1 - 3x}\right) = \frac{-3x^2 + 2x + 3}{(1 - 3x)^2}\] So the derivative of \(f(x) = \frac{x^2 + 1}{1 - 3x}\) is: \[f'(x) = \frac{-3x^2 + 2x + 3}{(1 - 3x)^2}\]

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Most popular questions from this chapter

Chapter 52

Find the derivative of: $\mathrm{f}(\mathrm{x})=\left[\left\\{\left(\mathrm{x}^{2}+1\right) \sqrt{\left. \left.\left(\mathrm{x}^{2}-1\right)\right\\} /\\{3 \mathrm{x}+2\\}\right]}\right.\right.$
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