3.3 Assess Your Understanding

Concepts and Vocabulary

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\(\dfrac{d}{dx}\ln x\)=_______.

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True or False \(\dfrac{d}{dx}x^{e}=e\,x^{e-1}.\)

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True or False \(\dfrac{d}{dx}\ln [x\sin ^{2}x ] =\frac{d}{dx} \ln x\cdot \dfrac{d}{dx}\ln \sin ^{2}x\).

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True or False \(\dfrac{d}{dx}\ln \pi =\dfrac{1}{\pi }\).

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\(\dfrac{d}{dx}\ln \vert x\vert =\) _______ for all \(x≠ 0\).

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\(\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right) ^{n}=\) _______.

Skill Building

In Problems 7–44, find \(y^\prime\).

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\(y=5\ln x\)

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\(y=-3\ln x\)

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\(y=\log _{2}u\)

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\(y=\log _{3}u\)

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\(y=(\cos x)(\ln x)\)

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\(y=(\sin x)(\ln x)\)

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\(y=\ln (3x)\)

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\(y=\ln \dfrac{x}{2}\)

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\(y=\ln (e^{t}-e^{-t})\)

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\(y=\ln (e^{at}+e^{-at})\)

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\(y=x\ln (x^{2}+4)\)

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\(y=x\ln (x^{2}+5x+1)\)

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\(y=v\ln \sqrt{v^{2}+1}\)

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\(y=v\ln \sqrt[3]{3v+1}\)

229

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\(y=\dfrac{1}{2}\ln \dfrac{1+x}{1-x}\)

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\(y=\dfrac{1}{2}\ln \dfrac{1+x^{2}}{1-x^{2}}\)

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\(y=\ln (\ln x)\)

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\(y=\ln \left( \ln \dfrac{1}{x}\right)\)

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\(y=\ln \dfrac{x}{\sqrt{x^{2}+1}}\)

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\(y=\ln \dfrac{4x^{3}}{\sqrt{x^{2}+4}}\)

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\(y=\ln \dfrac{(x^{2}+1)^{2}}{x\sqrt{x^{2}-1}}\)

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\(y=\ln \dfrac{x\sqrt{3x-1}}{(x^{2}+1)^{3}}\)

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\(y=\ln (\sin \theta )\)

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\(y=\ln (\cos \theta )\)

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\(y=\ln (x+\sqrt{x^{2}+4})\)

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\(y=\ln (\sqrt{x+1}+\sqrt{x})\)

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\(y=\log _{2}(1+x^{2})\)

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\(y=\log _{2}(x^{2}-1)\)

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\(y=\tan^{-1}(\ln x)\)

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\(y=\sin ^{-1}(\ln x)\)

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\(y=\ln (\tan ^{-1}t)\)

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\(y=\ln (\sin ^{-1}t)\)

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\(y=(\ln x)^{1/2}\)

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\(y=(\ln x)^{-1/2}\)

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\(y=\sin (\ln \theta)\)

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\(y=\cos (\ln \theta)\)

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\(y=x \ln \sqrt{\cos (2x) }\)

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\(y=x^{2}\ln \sqrt{\sin (2x) }\)

In Problems 45–50, use implicit differentiation to find \(y^\prime =\dfrac{dy}{dx}.\)

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\(x\ln y+y\ln x=2\)

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\(\dfrac{\ln y}{x}+\dfrac{\ln x}{y}=2\)

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\(\ln (x^{2}+y^{2})=x+y\)

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\(\ln (x^{2}-y^{2})=x-y\)

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\(\ln \dfrac{y}{x} =y\)

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\(\ln \dfrac{y}{x}-\ln \dfrac{x}{y}=1\)

In Problems 51–72, use logarithmic differentiation to find \(y^\prime\). Assume that the variable is restricted so that all arguments of logarithm functions are positive.

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\(y=(x^{2}+1)^{2}(2x^{3}-1)^{4}\)

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\(y=(3x^{2}+4)^{3}(x^{2}+1)^{4}\)

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\(y=\dfrac{x^{2}(x^{3}+1)}{\sqrt{x^{2}+1}}\)

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\(y=\dfrac{\sqrt{x}(x^{3}+2)^{2}}{\sqrt[3]{3x+4}}\)

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\(y=\dfrac{x\cos x}{(x^{2}+1)^{3}\sin x}\)

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\(y=\dfrac{x\sin x}{( 1+e^{x}) ^{3}\cos x}\)

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\(y=(3x)^{x}\)

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\(y=(x-1)^{x}\)

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\(y=x^{\ln x}\)

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\(y=(2x) ^{\ln x}\)

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\(y=x^{x^{2}}\)

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\(y=(3x)^{\sqrt{x}}\)

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\(y=x^{{e}^{{x}}}\)

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\(y=(x^{2}+1)^{{e}^{{x}}}\)

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\(y=x^{\sin x}\)

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\(y=x^{\cos x}\)

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\(y=(\sin x)^{x}\)

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\(y=(\cos x)^{x}\)

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\(y=(\sin x)^{\cos x}\)

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\(y=(\sin x)^{\tan x}\)

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\(x^{y}=4\)

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\(y^{x}=10\)

In Problems 73–76, find an equation of the tangent line to the graph of \(y=f(x)\) at the given point.

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\(y=\ln (5x)\) at \(\left(\dfrac{1}{5},0\right)\)

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\(y=x\ln x\) at \((1,0)\)

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\(y=\dfrac{x^{2}\sqrt{3x-2}}{( x-1) ^{2}}\) at \((2, 8)\)

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\(y=\dfrac{x (\sqrt[3]{x} +1) ^{2}}{\sqrt{x+1}}\) at \((8,24)\)

In Problems 77–80, express each limit in terms of \(e.\)

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\(\lim\limits_{n \rightarrow \infty}\left( 1+\dfrac{1}{n}\right) ^{2n}\)

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\(\lim\limits_{n \rightarrow \infty}\left(1+\dfrac{1}{n}\right)^{n/2}\)

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\(\lim\limits_{n \rightarrow \infty}\left(1+\dfrac{1}{3n}\right)^{n}\)

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\(\lim\limits_{n \rightarrow \infty}\left(1+\dfrac{4}{n}\right) ^{n}\)

Applications and Extensions

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Find \(\dfrac{d^{10}}{dx^{10}}(x^{9}\ln x)\).

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If \(f(x)=\ln (x-1)\), find \(f^{(n)}(x)\).

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If \(y=\ln (x^{2}+y^{2})\), find the value of \(\dfrac{dy}{dx}\) at the point \((1,0)\).

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If \(f(x) =\tan \left( \ln x-\dfrac{1}{\ln x} \right)\), find \(f^\prime (e)\).

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Find \(y^\prime\) if \(y=x^{x},\) \(x > 0\), by using \(y=x^{x}=e^{\ln x^{x}}\) and the Chain Rule.

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If \(y=\ln (kx)\), where \(x > 0\) and \(k > 0\) is a constant, show that \(y^\prime =\dfrac{1}{x}.\)

In Problems 87 and 88, find \(y^\prime\). Assume that \(a\) is a constant.

Question

\(y=x\tan ^{-1}\dfrac{x}{a}-\dfrac{1}{2}a\ln (x^{2}+a^{2}),\quad a≠ 0\)

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\(y=x\sin ^{-1}\dfrac{x}{a}+a\ln \sqrt{a^{2}-x^{2}}\), \(\vert a \vert > \vert x\vert\), \(a≠ 0\)

Continuously Compounded Interest In Problems 89 and 90, use the following discussion:

Suppose an initial investment, called the principal \(P\), earns an annual rate of interest \(r\), which is compounded \(n\) times per year. The interest earned on the principal \(P\) in the first compounding period is \(P\left( \dfrac{r}{n}\right)\), and the resulting amount \(A\) of the investment after one compounding period is \(A=P+P \left(\dfrac{r}{n}\right) =P\left( 1+\dfrac{r}{n}\right)\). After \(k\) compounding periods, the amount \(A\) of the investment is \(A=P\left(1+\dfrac{r}{n}\right)^{k}\). Since in \(t\) years there are \(nt\) compounding periods, the amount \(A\) after \(t\) years is \[ A=P\left( 1+\dfrac{r}{n}\right)^{nt} \] When interest is compounded so that after \(t\) years the accumulated amount is \(A=\) \(\lim\limits_{n\rightarrow \infty }P\left( 1+\dfrac{r}{n}\right) ^{nt}\), the interest is said to be compounded continuously.

Question

  1. Show that if the annual rate of interest \(r\) is compounded continuously, then the amount \(A\) after \(t\) years is \(A={\it Pe}^{rt}\), where \(P\) is the initial investment.
  2. If an initial investment of \(P=$5000\) earns 2% interest compounded continuously, how much is the investment worth after \(10\) years?
  3. How long does it take an investment of $10,000 to double if it is invested at 2.4% compounded continuously?
  4. Show that the rate of change of \(A\) with respect to \(t\) when the interest rate \(r\) is compounded continuously is \(\dfrac{dA}{dt}=rA.\)

230

Question

A bank offers a certificate of deposit (CD) that matures in \( 10\) years with a rate of interest of 3% compounded continuously. (See Problem 89.) Suppose you purchase such a CD for $2000 in your IRA.

  1. Write an equation that gives the amount \(A\) in the CD as a function of time \(t\) in years.
  2. How much is the CD worth at maturity?
  3. What is the rate of change of the amount \(A\) at \(t=3?\) At \(t=5?\) At \( t=8?\)
  4. Explain the results found in (c).

Question

Sound Level of a Leaf Blower The loudness \(L\), measured in decibels (dB), of a sound of intensity \(I\) is defined as \(L(x)=10\log \dfrac{ I(x)}{I_{0}}\), where \(x\) is the distance in meters from the source of the sound and \(I_{0}=10^{-12}\) W/m\(^{2}\) is the least intense sound that a human ear can detect. The intensity \(I\) is defined as the power \(P\) of the sound wave divided by the area \(A\) on which it falls. If the wave spreads out uniformly in all directions, that is, if it is spherical, the surface area is \(A(x)=4\pi x^{2}\)m\(^{2}\), and \(I(x)= \dfrac{P}{4\pi x^{2}}\)W/m\(^{2}\).

  1. If you are 2.0 m from a noisy leaf blower and are walking away from it, at what rate is the loudness \(L\) changing with respect to distance \(x\)?
  2. Interpret the sign of your answer.

Question

Show that \(\ln x+\ln y=2x\) is equivalent to \(xy=e^{2x}.\) Use this equation to find \(y^\prime\). Compare this result to the solution found in Example 1(c).

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If \(\ln T=kt\), where \(k\) is a constant, show that \(\dfrac{dT}{dt}=kT\).

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Graph \(y=\left( 1+\dfrac{1}{x}\right) ^{x}\) and \(y=e\) on the same set of axes. Explain how the graph supports the fact that \( \lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right) ^{n}=e\).

Question

Power Rule for Functions Show that if \(u\) is a function of \(x\) that is differentiable and \(a\) is a real number, then \[ \dfrac{d}{dx}[u( x) ] ^{a}=a[ u( x) ] ^{a-1}u^\prime (x) \] provided \(u^{a}\) and \(u^{a-1}\) are defined. [\(Hint:\) Let \(\vert y\vert =\vert [u( x)] ^{a}\vert\) and use logarithmic differentiation.]

Question

Show that the tangent lines to the graphs of the family of parabolas \(f( x) =-\dfrac{1}{2}x^{2}+k\) are always perpendicular to the tangent lines to the graphs of the family of natural logarithms \(g(x) =\ln (bx) +c\), where \(b > 0\), \(k\), and \(c\) are constants.

Source: Mathematics students at Millikin University, Decatur, Illinois.

Challenge Problems

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Show that \(2x-\ln (3+6e^{x}+3e^{2x})=C-2\ln (1+e^{-x})\) for some constant \(C\).

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If \(f\) and \(g\) are differentiable functions, and if \(f(x) > 0\), show that \[ \frac{d}{dx}f(x)^{g(x)}=g(x)f(x)^{g(x)-1}f^\prime ( x) +f( x) ^{g(x)}[\ln f(x)]g^\prime (x) \]