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Chapter 4: Problem 3

Differentiate (a) \(y=x^{4} \ln x\) (b) \(y=\mathrm{e}^{x^{2}}\) (c) \(y=\frac{\ln x}{x+2}\)

Short Answer

Expert verified
a) \( y' = 4x^3 \ln x + x^3 \) b) \( y' = 2x \mathrm{e}^{x^2} \) c) \( y' = \frac{1 + 2x - x\ln x}{(x + 2)^2} \)

Step by step solution

01

Part (a): Apply Product Rule

For the function \(y = x^4 \ln x\), use the product rule: \((uv)' = u'v + uv'\). Let \(u = x^4\) and \(v = \ln x\).
02

Differentiate \(u = x^4\)

Find the derivative of \(u\). We have \(u' = 4x^3\).
03

Differentiate \(v = \ln x\)

Find the derivative of \(v\). We have \(v' = \frac{1}{x}\).
04

Combine Derivatives

Use the product rule: \(y' = u'v + uv' = 4x^3 \ln x + x^4 \frac{1}{x}\). Simplify to \(y' = 4x^3 \ln x + x^3\).
05

Part (b): Differentiate Using Chain Rule

For the function \(y = \mathrm{e}^{x^2}\), use the chain rule: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\). Let \(g(x) = x^2\) and \(f(u) = \mathrm{e}^u\).
06

Differentiate \(g(x) = x^2\)

Find the derivative of \(g(x)\). We have \(g'(x) = 2x\).
07

Differentiate \(f(u) = \mathrm{e}^u\)

The derivative of \(f(u) = \mathrm{e}^u\) is \(f'(u) = \mathrm{e}^u\).
08

Combine Derivatives

Using the chain rule: \(y' = f'(g(x)) \cdot g'(x) = \mathrm{e}^{x^2} \cdot 2x = 2x \mathrm{e}^{x^2}\).
09

Part (c): Apply Quotient Rule

For the function \(y = \frac{\ln x}{x + 2}\), use the quotient rule: \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\). Let \(u = \ln x\) and \(v = x + 2\).
10

Differentiate \(u = \ln x\)

Find the derivative of \(u\). We have \(u' = \frac{1}{x}\).
11

Differentiate \(v = x + 2\)

Find the derivative of \(v\). We have \(v' = 1\).
12

Combine Derivatives

Using the quotient rule: \(y' = \frac{u'v - uv'}{v^2} = \frac{\left(\frac{1}{x}\right)(x + 2) - (\ln x)(1)}{(x + 2)^2} = \frac{1 + 2x - x\ln x}{(x + 2)^2}\).

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Key Concepts

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

product rule
The product rule is a fundamental concept in calculus used to differentiate functions that are products of two or more functions. It states that if you have two functions, say \(u(x)\) and \(v(x)\), their product \(w(x)\) can be differentiated as:
\[ w'(x) = u'(x)v(x) + u(x)v'(x) \]
Let's see how it works in practice with the example \(y = x^{4} \ln x\):
  • First, identify the two separate functions: \(u(x) = x^4\) and \(v(x) = \ln x\).
  • Next, differentiate each function:
    \(u'(x) = 4x^3\) and \(v'(x) = \frac{1}{x}\).
  • Finally, plug these derivatives into the product rule formula:
    \(y' = 4x^3 \ln x + x^4 \frac{1}{x} \).
    This simplifies to
    \(y' = 4x^3 \ln x + x^3\).
The product rule is handy when dealing with multiplicative functions, making it easier to manage the differentiation process efficiently.
chain rule
The chain rule is another vital differentiation method in calculus. It helps us differentiate composite functions, which are functions within other functions. The chain rule states:
If you have a function \(y = f(g(x))\), then the derivative \(y'\) is:
\[ y' = f'(g(x)) \, g'(x) \]
Let's apply the chain rule to the function \(y = \mathrm{e}^{x^2}\):
  • First, identify the outer function \(f(u) = \mathrm{e}^u\) and the inner function \(g(x) = x^2\).
  • Differentiate both functions: \( f'(u) = \mathrm{e}^u \, and \ g'(x) = 2x \).
  • Combine these using the chain rule formula:
    \(y' = \mathrm{e}^{x^2} \, 2x\).
  • This result simplifies directly to \(y' = 2x \mathrm{e}^{x^2}\).
The chain rule is especially useful for functions nested within other functions, providing a systematic way to tackle complex differentiation problems.
quotient rule
The quotient rule assists with differentiating ratios of two functions. This is particularly important when the functions are difficult to split apart. The quotient rule is given by:
If you have a function \(y = \frac{u(x)}{v(x)}\), then its derivative \(y'\) is:
\[ y' = \frac{u'(x)v(x) - u(x)v'(x)}{v^2(x)} \]
Let's use this rule on the function \(y = \frac{\ln x}{x + 2}\):
  • Identify the numerator \(u = \ln x\) and the denominator \(v = x + 2\).
  • Differentiate both:
    \(u'(x) = \frac{1}{x}\) and \(v'(x) = 1\).
  • Apply the quotient rule:
    \(y' = \frac{(\frac{1}{x})(x + 2) - (\ln x)(1)}{(x+2)^2}\).
  • Simplify the expression to get \(y' = \frac{1 + 2x - x \ln x}{(x+2)^2}\).
The quotient rule is essential when dealing with functions as fractions, ensuring accurate and efficient differentiation.

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Most popular questions from this chapter

The economic order quantity, EOQ, is used in cost accounting to minimize the total cost, TC, to order and carry a firm's stock over the period of a year. The annual cost of placing orders, \(A C O\), is given by $$ \mathrm{ACO}=\frac{(\mathrm{ARU})(\mathrm{CO})}{\mathrm{EOQ}} $$ where $$ \begin{aligned} \mathrm{ARU} &=\text { annual required units } \\ \mathrm{CO} &=\text { cost per order } \end{aligned} $$ The annual carrying cost, ACC, is given by $$ \mathrm{ACC}=(\mathrm{CU})(\mathrm{CC}) \frac{(\mathrm{EOQ})}{2} $$ where $$ \begin{aligned} &\mathrm{CU}=\text { cost per unit } \\ &\mathrm{CC}=\text { carrying } \mathrm{cost} \end{aligned} $$ and \((E O Q) / 2\) provides an estimate of the average number of units in stock at any given time of the year. Assuming that \(\mathrm{ARU}, \mathrm{CO}, \mathrm{CU}\) and \(\mathrm{CC}\) are all constant, show that the total cost $$ \mathrm{TC}=\mathrm{ACO}+\mathrm{ACC} $$ is minimized when $$ \mathrm{EOQ}=\sqrt{\frac{2(\mathrm{ARU})(\mathrm{CO})}{(\mathrm{CU})(\mathrm{CC})}} $$

Find the output needed to maximize profit given that the total cost and total revenue functions are \(\mathrm{TC}=2 Q \quad\) and \(\quad \mathrm{TR}=100 \ln (Q+1)\) respectively.

If the supply equation is $$ Q=7+0.1 P+0.004 P^{2} $$ find the price elasticity of supply if the current price is 80 . (a) Is supply elastic, inelastic or unit elastic at this price? (b) Estimate the percentage change in supply if the price rises by \(5 \%\).

If the demand equation is $$ P=200-40 \ln (Q+1) $$ calculate the price elasticity of demand when \(Q=20\).

Differentiate (a) \(y=x(x-3)^{4}\) (b) \(y=x \sqrt{(2 x-3)}\) (c) \(y=\frac{x}{x+5}\) (d) \(y=\frac{x}{x^{2}+1}\)
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