Homework 4, due July 13 at beginning of class.

  1. Show that $\log_{a}x=\frac{\log_{b}x}{\log_{b}a}$. (Hint: take $a$ to the power of boths sides. On the left side of the equation, you will just get $x$, and on the right side you'll get $a$ to that fraction there, but now replace $a$ with $b^{\log_{b} a}$).
    Answer: Note that since $a=b^{\log_{b}a}$, \[ a^{\frac{\log_{b}x}{\log_{b}a}}=(b^{\log_{b}a})^{\frac{\log_{b}x}{\log_{b}a}}=b^{(\log_{b}a)\frac{\log_{b}x}{\log_{b}a}}=b^{\log_{b}x}=x,\] but $a^{\log_{a}x}=x$ as well, thus \[a^{\log_{a}x}=a^{\frac{\log_{b}x}{\log_{b}a}}\] and hence \[\log_{a}x=\frac{\log_{b}x}{\log_{b}a}.\]

     

  2. Differentiate the following functions:
    1. $\ln \cos x$
      Answer: This is chain rule, $\ln \cos x =f(g(x))$ where $f(x)=\ln x$ and $g(x)=\cos x$, so $f'(x)=\frac{1}{x}$ and $g'(x)=-\sin x$, hence \[(\ln\cos x)'=(f(g(x))'=f'(g(x))g'(x)=\frac{1}{\cos x}(-\sin x)=-\tan x.\]
    2. $x\sin \ln x$ First, we do product rule, \[(x\sin \ln x)'=(x')\sin \ln x+x(\sin \ln x)'.\] We know $(x)'=1$ by the power rule. For $(\sin \ln x)'$, we use chain rule, the outside function being $\sin x$ (with derivative $\cos x$) and inside function $\ln x$ (with derivative $\frac{1}{x}$,) so \[(\sin \ln x)'=(\cos\ln x)\frac{1}{x}=\frac{\cos\ln x}{x},\] hence \[(x\sin \ln x)'=(x')\sin \ln x+x(\sin \ln x)'=1\cdot \sin \ln x+x\left(\frac{\cos \ln x}{x}\right)=\sin\ln x+\cos\ln x.\]
    3. $e^{x^{2}}$
      Answer: We use chain rule: $e^{x^{2}}=f(g(x))$ where $f(x)=e^{x}$ and $g(x)=x^{2}$, so $f'(x)=e^{x}$ and $g'(x)=2x$, hence \[(e^{x^{2}})'=(f(g(x))'=f'(g(x))g'(x)=e^{x^{2}}(2x)=2xe^{x^{2}}.\]
    4. $\ln\sqrt{1-x^{2}}$
    5. $e^{\sin e^{2x}}$
    6. $x\ln x-x$
      Answer: \[(x\ln x-x)'=(x\ln x)'-(x)'=(1\cdot \ln x+ x(\frac{1}{x}))-(1)=\ln x+1-1=\ln x.\] In the second equality, we used product rule on $x\ln x$.
    7. $\log_{a} x$
      Answer: Note that by problem 1, with $b=e$, \[\log_{a}x=\frac{\log_{e}x}{\log_{e}a}=\frac{\ln x}{\ln a},\] hence \[(\log_{a}x)'=\left(\frac{\ln x}{\ln a}\right)'=\frac{1}{\ln a}(\ln x)'=\frac{1}{\ln a}\frac{1}{x}=\frac{1}{x\ln a}.\]
    8. $e^{\frac{1}{\ln x}}$
      Answer: Chain rule gives \[(e^{\frac{1}{\ln x}})'=e^{\frac{1}{\ln x}}\left(\frac{1}{\ln x}\right)'.\] To compute this last derivative, we use quotient rule, \[\left(\frac{1}{\ln x}\right)'=\frac{(1)'\ln x-1\cdot(\ln x)'}{(\ln x)^{2}}=\frac{0\cdot \ln x-1\cdot\frac{1}{x}}{(\ln x)^{2}}=\frac{1}{x(\ln x)^{2}}\] hence \[(e^{\frac{1}{\ln x}})'=e^{\frac{1}{\ln x}}\left(\frac{1}{\ln x}\right)'=e^{\frac{1}{\ln x}} \frac{1}{x(\ln x)^{2}}\]

     

  3. Compute the following derivatives:
    1. $(\sin x)^{\sin x}$
      Answer: Set $f(x)=(\sin x)^{\sin x}$ and take $\ln $ of both sides of this to get \[\ln f(x)=\ln(\sin x)^{\sin x}=(\sin x)\ln \sin x,\] and differentiating both sides gives \[\frac{f'(x)}{f(x)}=(\cos x)\ln \sin x+\sin x \frac{1}{\cos x}\] and multiplying both sides by $f(x)=(\sin x)^{\sin x}$ gives \[f'(x)=(\sin x)^{\sin x}\left((\cos x)\ln \sin x+\frac{\sin x}{\cos x}\right)\]
    2. $(x^{2}+1)^{e^{5x}}$
      Answer: Again, set $f(x)=(x^{2}+1)^{e^{5x}}$, take $\ln$ of both sides, \[\ln f(x)=\ln(x^{2}+1)^{e^{5x}}=e^{5x}\ln(x^{2}+1),\] and differentiating both sides gives \[\frac{f'(x)}{f(x)}=5e^{5x}\ln(x^{2}+1)+e^{5x}\frac{1}{x^{2}+1}2x\] and multiplying both sides by $f(x)$ gives \[f'(x)=(x^{2}+1)^{e^{5x}}\left( 5e^{5x}\ln)x^{2}+1)+e^{5x}\frac{2x}{x^{2}+1}\right).\]
    3. $\frac{x^{\sin^{2}x }}{1+x^{\cos^{2}x}}$
    4. $x^{x^{x}}$
      Answer: Let $f(x)=x^{x^{x}}$. By what we did in class, \[(x^{x})'=x^{x}(\ln x+1),\] So if we take $\ln f(x)$, \[\ln f(x)=\ln x^{x^{x}}=x^{x}\ln x,\] and differentiating \[\frac{f'(x)}{f(x)}=(x^{x})'\ln x+x^{x}(\ln x)'=x^{x}(\ln x+1)\ln x+x^{x}\frac{1}{x}.\]

     

  4. Compute $(f^{-1})'(f(c))$ where $f$ and $c$ are:
    1. $f(x)= 2x+\cos x$, $c=\frac{\pi}{2}$
      Answer: For all these problems, we first obseve that for any $c$, \[(f^{-1})'(f(c))=\frac{1}{f'(f^{-1}(f(c)))}=\frac{1}{f'(c)},\] so for this particular problem, $f'(x)=2-\sin x$, so \[(f^{-1})'(f(2))=\frac{1}{f'(2)}=\frac{1}{2-\sin 2}.\]
    2. $f(x)=\frac{e^{x}-e^{-x}}{2}$, $c=\ln 2$
      Answer: For this one, note $f'(x)=\frac{e^{x}+e^{-x}}{2}$, hence \[(f^{-1})'(f(\ln 2))=\frac{1}{f'(\ln 2)}=\frac{1}{\frac{e^{\ln 2}+e^{-\ln 2}}{2}}=\frac{2}{2+\frac{1}{2}}=\frac{4}{5}.\]

     

  5. Compute $(\arctan x)'$ (Hint: you will need the equation $\sec^{2} x=1+\tan^{2}x$.
    Answer: We use the inverse formula (since $\arctan x$ is the inverse of $\tan x$), \[(\arctan x)'=\frac{1}{\tan '(\arctan x)}.\] So we need to compute the derivative of $\tan$, \[(\tan x)'=\left(\frac{\sin x}{\cos x}\right)'=\frac{(\cos x)(\cos x)-(\sin x)(-\sin x)}{\cos^{2}x}=\frac{\cos^{2} x+\sin^{2} x}{\cos^{2}x}=\frac{1}{\cos^{2} x}=\sec^{2}x.\] Hence, using our earlier formula, we get \[(\arctan x)'=\frac{1}{\sec^{2}(\arctan x)}=\frac{1}{1+\tan^{2}(\arctan x)}=\frac{1}{1+(\tan\arctan x)^{2}}=\frac{1}{1+x^{2}}.\] In the second equation, we use the formula in the hint to convert $\sec^{2}$ to $1+\tan^{2}$.

     

  6. Compute $(\arccos x)'$ (Hint: you will need the equation $\sin(x)=\sqrt{1-\cos^{2}(x)}$.

     

  7. Solve for $\frac{dy}{dx}$ at the given points:
    1. $x^{2}-y^{4}=x^{3}+y^{5}$ at $(x,y)=(-1,1)$
      Answer: We first differentiate both sides to get \[2x-4y^{3}y'=3x^{2}+5y^{4}y',\] and plug in $(x,y)=(-1,1)$ to get \[2(-1)-4(1)^{3}y'=3(-1)^{2}+5(1)^{4}y'.\] Simplifying, we get \[-2-4y'=3+5y',\] and solving for $y'$, we get $y'=\frac{-5}{9}$.
    2. $\cos(x+y) -xy=1+\frac{\pi^{2}}{16}$ at $(x,y)=(\frac{\pi}{4},-\frac{\pi}{4})$,
    3. $\frac{x}{y}=\sin \frac{\pi xy}{2}$ at $(x,y)=(1,1)$
      Answer: Differentate both sides, \[\frac{y-xy'}{y^{2}}=(\cos\frac{\pi xy}{2})\frac{\pi}{2}(y+xy')\] and plugging in $(x,y)=(1,1)$ gives \[1-y'=0,\] so $y'=1.

     

  8. Show that $\lim_{h\rightarrow 0} (1+h)^{\frac{1}{h}}=e$ (Hint: Take $\ln$ of the quantity in the limit and note that, by the definition of the derivative, $\frac{1}{x}=(\ln x)'=\lim_{h\rightarrow 0}\frac{\ln(x+h)-\ln x}{h}$, so $\frac{1}{1}=\lim_{h\rightarrow 0}\frac{\ln(1+h)-\ln 1}{h}=\lim_{h\rightarrow 0}\frac{\ln(1+h)}{h}$.
    Answer: Note that \[(\ln x)'=\lim_{h\rightarrow 0}\frac{\ln (x+h)-\ln x}{h}$, \] and since we know $(\ln x)'=\frac{1}{x}$, we now know \[\frac{1}{x}=\lim_{h\rightarrow 0}\frac{\ln(x+h)-\ln x}{h}.\] In the special case that $x=1$, this gives \[1=\lim_{h\rightarrow 0}\frac{\ln(1+h)-\ln 1}{h}=\lim_{h\rightarrow 0}\frac{\ln (1+h)}{h}=\lim_{h\rightarrow 0}\frac{1}{h}\ln(1+h)=\lim_{h\rightarrow 0}\ln(1+h)^{\frac{1}{h}},\] and exponentiating both ends of this string of equations gives \[e^{1}=e=e^{\lim_{h\rightarrow 0}\ln(1+h)^{\frac{1}{h}}}=\lim_{h\rightarrow 0}e^{\ln(1+h)^{\frac{1}{h}}}=\lim_{h\rightarrow 0}(1+h)^{\frac{1}{h}}$,\] where we pulled $e$ inside the limit because $e^{x}$ is a continuous function.

     

  9. Suppose the radius of a tube of height 1 is expanding, and suppose that when the radius is $2$, the volume is expanding at $\pi$ units a second. How fast is the radius increasing at this point in time? (Recall the volume of a cylinder is $\pi r^{2}h$ where $r$ is the radius and $h$ is the height.)
    Answer: Let's call the time when the radius is two $t_{0}$, so $r=r(t)$ is a function that is increasing in time, and $r(t_{0})=2$. The volume $V(t)$ is also a function of time, and is equal to \[V(t)=\pi r(t)^{2}h=\pi r^{2},\] note that the $h$ disappeared since it is just $1$ in this problem. We know what the values of $V'(t)$ and $r(t)$ are when $t=t_{0}$ (they are $\pi$ and $2$ respectively), but if we compute the formula for the derive of $V$, we will get an expression that also involves $r'(t)$: \[V'(t)=(\pi r(t)^{2})'=\pi (2r(t)r'(t))=2\pi r(t)r'(t).\] Evaluating this at $t=t_{0}$, we get \[\pi = V'(t_{0})=2\pi (2) r'(t_{0})=4\pi r'(t_{0}).\] Dividing both sides by $4\pi$ gives \[r'(t_{0})=\frac{1}{4}.\]

     

  10. Suppose that there is a cone starting out with a base of radius $1$ and height $4$, and whose base is increasing but its volume remains constant. Suppose when the radius is $2$ its radius is increasing at a rate of $2$ units per second. How fast is its height decreasing (i.e. what is the derivative of its height) when $r=2$? (Hint: the volume of a cone is $\frac{1}{3}\pi r^{2}h$), solve for $h$ in terms of $r$).
    Answer: Note that $h(t)$ and $r(t)$ are both functions that are changing in time, but the volume of the cone $V=\frac{1}{3}\pi r(t)^{2}h(t)$ is constant, and is equal to whatever it was at the start, i.e. when the radius is 1 and height is 4, which gives $V=\frac{1}{3}\pi 1^{2}\cdot 4=\frac{4}{3}\pi$. We want to compute $h'(t)$ when $t$ is the time that the radius is 2 (call this time $t_{0}$), so lets solve for $h(t)$, \[h(t)=\frac{3V}{\pi r(t)^{2}}=\frac{3\frac{4}{3}\pi}{\pi r(t)^{2}}=\frac{4}{r(t)^{2}},\] and differentiating gives \[h'(t)=-\frac{8}{r(t)^{3}}r'(t),\] and at $t=t_{0}$, we know that $r(t_{0})=2$ and $r'(t_{0})=2$, so \[h'(t_{0})=-\frac{8}{2^{3}}2=-2.\]