/*! 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 55 Find \(\frac{d y}{d x}\) $$ ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 55

Find \(\frac{d y}{d x}\) $$ y=\pi x^{\pi}+\ln x^{2} $$

Short Answer

Expert verified
\(\frac{dy}{dx} = \pi^2 x^{\pi - 1} + \frac{2}{x}\)."

Step by step solution

01

Differentiate the First Term

The first term is \(\pi x^{\pi}\). To differentiate this with respect to \(x\), apply the power rule: \(\frac{d}{dx}x^n = n x^{n-1}\), and treat \(\pi\) as a constant. Thus, \(\frac{d}{dx}(\pi x^{\pi}) = \pi \cdot \pi \cdot x^{\pi - 1} = \pi^2 x^{\pi - 1}\).
02

Differentiate the Second Term

The second term is \(\ln x^2\). Using the chain rule \(\frac{d}{dx}(\ln u) = \frac{1}{u}\frac{du}{dx}\), first identify \(u = x^2\), which gives \(\frac{du}{dx} = 2x\). So, \(\frac{d}{dx}(\ln x^2) = \frac{1}{x^2} \cdot 2x = \frac{2}{x}\).
03

Summing Up the Derivatives

Combine the results from Step 1 and Step 2. The derivative \(\frac{dy}{dx}\) is the sum of the derivatives of the separate terms: \[\frac{dy}{dx} = \pi^2 x^{\pi - 1} + \frac{2}{x}\].

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Key Concepts

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

Power Rule
The power rule is a fundamental tool in calculus for finding the derivative of a function of the form \(x^n\), where \(n\) is a constant. It's quite simple:
  • The rule states that if you have \(f(x) = x^n\), then the derivative \(f'(x) = nx^{n-1}\).
  • This means you multiply the exponent by the coefficient (if there's one), then reduce the exponent by one.
  • In our exercise, we applied the power rule to \(\pi x^{\pi}\). Treat \(\pi\) as a constant multiplier, similar to how you would treat any other number, like 5 or 10.
  • Thus, the derivative becomes \(\pi \cdot \pi \cdot x^{\pi-1} = \pi^2 x^{\pi-1}\).
The power rule is incredibly useful for differentiating polynomial functions and simplifying expressions, as it allows you to calculate derivatives quickly and efficiently. Always remember: decrease the power by one and multiply it with its original value.
Chain Rule
The chain rule is a method for finding the derivative of composite functions. A composite function is when you take a function and place another function inside it, like \(f(g(x))\).
  • The chain rule formula is: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\).
  • In simpler terms, it means you first differentiate the outer function, leaving the inside function unchanged, then multiply it by the derivative of the inside function.
  • In our problem, we used the chain rule on \(\ln(x^2)\). Here, your inside function is \(x^2\), and the outside function is \(\ln u\), where \(u = x^2\).
  • Differentiate \(\ln u\) as \(\frac{1}{u}\) and \(x^2\) as \(2x\). Thus, applying the chain rule gives us \(\frac{1}{x^2} \cdot 2x = \frac{2}{x}\).
This approach becomes crucial in more complex functions, allowing you to systematically tackle derivatives of functions within functions. Always identify your inner and outer functions correctly to apply the rule effectively.
Derivative of Logarithmic Function
Finding the derivative of a logarithmic function like \(\ln x\) involves its own special rules. Logarithmic functions and their derivatives are key components in calculus.
  • For a basic natural logarithm, \(\ln x\), the derivative is simply \(\frac{1}{x}\).
  • However, things become interesting when you have \(\ln\) of something more complex, say \(\ln(x^2)\), which we had in our problem.
  • Here, you apply not just the usual derivative rule of ln but also use the chain rule to incorporate the transformation \(x\) undergoes (which is into \(x^2\)).
  • The derivative of \(\ln(x^2)\) as derived is \(\frac{2}{x}\), since we had to differentiate both the outer function, \(\ln u\), and the inner function, \(x^2\), using the chain rule.
  • Recognizing when to use these derivative rules for logarithmic functions is crucial, especially because they pop up often in growth models, financial calculations, and more. With practice, handling them becomes an intuitive process.
This exercise showcases the interconnectedness of logarithmic differentiation and other calculus rules, reminding you that derivatives aren鈥檛 isolated procedures, but part of a larger mathematical toolkit.

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Most popular questions from this chapter

Medicine Poiseuille's law (see Clow and Urquhart \(^{20}\) ) states that the total resistance \(R\) to blood flow in a blood vessel of constant length \(l\) and radius \(r\) is given by \(R=\frac{a l}{r^{4}}\) where \(a\) is a positive constant. Find the rate of change of resistance with respect to the radius when \(r=3\).

Biology Talekar and coworkers \(^{91}\) showed that the number \(y\) of the eggs of \(O\). furnacalis per mung bean plant was approximated by \(y=f(x)=-57.40+63.77 x-\) \(20.83 x^{2}+2.36 x^{3},\) where \(x\) is the age of the mung bean plant and \(2 \leq x \leq 7\). Find \(f^{\prime}(x),\) and explain what this means. Use the quadratic formula to show that \(f^{\prime}(x)>0\) for \(2 \leq x \leq 7\). What does this say about the preference of O. furnacalis to laying eggs on older plants?

In Exercises 25 through \(32,\) use the quotient rule to find \(\frac{d y}{d u}\). $$ y=\frac{u^{3}-1}{u^{2}+3} $$

Transportation Costs Koo and coworkers \(^{89}\) used production cost data from the U.S. Department of Agriculture to estimate transportation costs for durum wheat. They found that the rail transportation cost \(R\) in dollars per ton from producing regions to durum mills was approximated by the equation \(R(D)=0.0143 D^{0.67743},\) where \(D\) is the distance in miles. Graph. Find values for \(R^{\prime}(D)\) at \(D=100,200\), \(400,\) and \(800 .\) Interpret what these numbers mean. What is happening? What does this mean in terms of economies of scale?

Find the derivative. \(e^{=x^{2}} \ln \left(x^{2}\right)\)
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