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Concepts and Vocabulary
ddxlnx=_______.
1x
True or False ddxxe=exe−1.
True
True or False ddxln[xsin2x]=ddxlnx⋅ddxlnsin2x.
False
True or False ddxlnπ=1π.
False
ddxln|x|= _______ for all x≠0.
1x
lim _______.
e
Skill Building
In Problems 7–44, find y^\prime.
y=5\ln x
y'=5/x
y=-3\ln x
y=\log _{2}u
y'={1\over u \ \ln 2}
y=\log _{3}u
y=(\cos x)(\ln x)
y'={1\over x}\cos x- \ \sin x \ \ln x
y=(\sin x)(\ln x)
y=\ln (3x)
y'=1/x
y=\ln \dfrac{x}{2}
y=\ln (e^{t}-e^{-t})
y'={e^t+e^{-t}\over e^t-e^{-t}}
y=\ln (e^{at}+e^{-at})
y=x\ln (x^{2}+4)
y'=\ln(x^2+4)+{2x^2\over x^2+4}
y=x\ln (x^{2}+5x+1)
y=v\ln \sqrt{v^{2}+1}
y'=\ln\big(\sqrt{v^2+1}\big)+{v^2\over v^2+1}
y=v\ln \sqrt[3]{3v+1}
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y=\dfrac{1}{2}\ln \dfrac{1+x}{1-x}
y'={1\over 1-x^2}
y=\dfrac{1}{2}\ln \dfrac{1+x^{2}}{1-x^{2}}
y=\ln (\ln x)
y'={1\over x \ \ln x}
y=\ln \left( \ln \dfrac{1}{x}\right)
y=\ln \dfrac{x}{\sqrt{x^{2}+1}}
y'={1\over x(x^2+1)}
y=\ln \dfrac{4x^{3}}{\sqrt{x^{2}+4}}
y=\ln \dfrac{(x^{2}+1)^{2}}{x\sqrt{x^{2}-1}}
y'=\dfrac{4x}{x^2+1} -\dfrac{1}{x} - \dfrac{x}{x^2-1}
y=\ln \dfrac{x\sqrt{3x-1}}{(x^{2}+1)^{3}}
y=\ln (\sin \theta )
y'=\cot\theta
y=\ln (\cos \theta )
y=\ln (x+\sqrt{x^{2}+4})
y'={1\over \sqrt{x^2+4}}
y=\ln (\sqrt{x+1}+\sqrt{x})
y=\log _{2}(1+x^{2})
y'={2x\over (1+x^2) \ \ln 2}
y=\log _{2}(x^{2}-1)
y=\tan^{-1}(\ln x)
y'={1\over x(1+(\ln x)^2)}
y=\sin ^{-1}(\ln x)
y=\ln (\tan ^{-1}t)
y'={1\over \tan^{-1} t}\cdot {1\over 1+t^2}
y=\ln (\sin ^{-1}t)
y=(\ln x)^{1/2}
y'={1\over 2x(\ln x)^{1/2}}
y=(\ln x)^{-1/2}
y=\sin (\ln \theta)
y'=\cos(\ln\theta)\cdot{1\over \theta}
y=\cos (\ln \theta)
y=x \ln \sqrt{\cos (2x) }
y'=\ln\sqrt{\cos(2x)}-x \ \tan(2x)
y=x^{2}\ln \sqrt{\sin (2x) }
In Problems 45–50, use implicit differentiation to find y^\prime =\dfrac{dy}{dx}.
x\ln y+y\ln x=2
{dy\over dx}=- \dfrac{y^2 +xy \ \ln y}{x^2 +xy \ \ln x}
\dfrac{\ln y}{x}+\dfrac{\ln x}{y}=2
\ln (x^{2}+y^{2})=x+y
{dy\over dx}={x^2+y^2-2x\over 2y-x^2-y^2}
\ln (x^{2}-y^{2})=x-y
\ln \dfrac{y}{x} =y
{dy\over dx}={y\over x(1-y)}
\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.
y=(x^{2}+1)^{2}(2x^{3}-1)^{4}
y'=(x^2+1)^2(2x^3-1)^4\left({4x\over x^2+1}+{24x^2\over 2x^3-1}\right)
y=(3x^{2}+4)^{3}(x^{2}+1)^{4}
y=\dfrac{x^{2}(x^{3}+1)}{\sqrt{x^{2}+1}}
y'={x^2(x^3+1)\over \sqrt{x^2+1}} \left({2\over x}+{3x^2\over x^3+1}-{x\over x^2+1}\right)
y=\dfrac{\sqrt{x}(x^{3}+2)^{2}}{\sqrt[3]{3x+4}}
y=\dfrac{x\cos x}{(x^{2}+1)^{3}\sin x}
y'={x \ \cos x\over (x^2+1)^3 \ \sin x}\left({1\over x}- \ \tan x-{6x\over x^2+1}-\ \cot x\right)
y=\dfrac{x\sin x}{( 1+e^{x}) ^{3}\cos x}
y=(3x)^{x}
y'=(3x)^x(\ln(3x)+1)
y=(x-1)^{x}
y=x^{\ln x}
y'=2x^{\ln x}{\ln x\over x}
y=(2x) ^{\ln x}
y=x^{x^{2}}
y'=x^{x^2}\left(2x\ln x+x\right)
y=(3x)^{\sqrt{x}}
y=x^{{e}^{{x}}}
y'=x^{e^x}\left(e^x\, \ln\, x+{e^x\over x}\right)
y=(x^{2}+1)^{{e}^{{x}}}
y=x^{\sin x}
y'=x^{\sin x}\left(\cos \,x\,\ln x+{\sin x\over x}\right)
y=x^{\cos x}
y=(\sin x)^{x}
y'=(\sin x)^x\left(\ln(\sin x)+x\cot x\right)
y=(\cos x)^{x}
y=(\sin x)^{\cos x}
y'=(\sin x)^{\cos x}\left(-\sin x\, \ln(\sin x)+\,\cos x\,\cot x\right)
y=(\sin x)^{\tan x}
x^{y}=4
y'={-y\over x\,\ln x}
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.
y=\ln (5x) at \left(\dfrac{1}{5},0\right)
y=5 x-1
y=x\ln x at (1,0)
y=\dfrac{x^{2}\sqrt{3x-2}}{( x-1) ^{2}} at (2, 8)
y=-5 x+18
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.
\lim\limits_{n \rightarrow \infty}\left( 1+\dfrac{1}{n}\right) ^{2n}
e^2
\lim\limits_{n \rightarrow \infty}\left(1+\dfrac{1}{n}\right)^{n/2}
\lim\limits_{n \rightarrow \infty}\left(1+\dfrac{1}{3n}\right)^{n}
e^{1/3}
\lim\limits_{n \rightarrow \infty}\left(1+\dfrac{4}{n}\right) ^{n}
Applications and Extensions
Find \dfrac{d^{10}}{dx^{10}}(x^{9}\ln x).
{d^{10}y\over dx^{10}}=\frac{362880}{x}
If f(x)=\ln (x-1), find f^{(n)}(x).
If y=\ln (x^{2}+y^{2}), find the value of \dfrac{dy}{dx} at the point (1,0).
{dy\over dx}=2
If f(x) =\tan \left( \ln x-\dfrac{1}{\ln x} \right), find f^\prime (e).
Find y^\prime if y=x^{x}, x > 0, by using y=x^{x}=e^{\ln x^{x}} and the Chain Rule.
y'=x^x(\ln\, x+1)
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.
y=x\tan ^{-1}\dfrac{x}{a}-\dfrac{1}{2}a\ln (x^{2}+a^{2}),\quad a≠ 0
y'=\tan^{-1}(x/a)
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.
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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.
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}.
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).
If \ln T=kt, where k is a constant, show that \dfrac{dT}{dt}=kT.
See Student Solutions Manual.
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.
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.]
See Student Solutions Manual.
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
Show that 2x-\ln (3+6e^{x}+3e^{2x})=C-2\ln (1+e^{-x}) for some constant C.
See Student Solutions Manual.
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)