Final Exam: problems and solutions

  1. Show that $(\ln x)'=\frac{1}{x}$ by using the fact $\ln x=f^{-1}(x)$, where $f(x)=e^{x}$.
    Answer: Here, we ust the above facts, the derivative formula for $(f^{-1}(x))'$, and the fact that $(e^{x})'=e^{x}$, to get \[(\ln x)'=(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}=\frac{1}{e^{f^{-1}(x)}}=\frac{1}{e^{\ln x}}=\frac{1}{x}.\]
  2. Use the Intermediate Value Theorem to show that there is $x\in [0,\frac{\pi}{2}]$ so that $\cos x= -\sin 3x$.
    Answer: Let $f(x)=\cos x+\sin 3x$. Note that \[f(\frac{\pi}{2})=\cos \frac{\pi}{2}+\sin \frac{3\pi}{2}=-1\leq 0 \leq f(0)=\cos 0-\sin 3\cdot 0=1,\] and so by the intermediate value theorem, there is $x\in [0,\frac{\pi}{2}]$ so that $f(x)=\cos x+\sin 3x=0$, i.e. so that $\cos x=-\sin 3x$.
  3. Compute the derivative of $\sqrt{x^{2}+1}$ using the limit definition of the derivative.
    Answer: \begin{align*} (\sqrt{x^{2}+1})' & = \lim_{h\rightarrow 0}\frac{\sqrt{(x+h)^{2}+1}-\sqrt{x^{2}+1}}{h} =\lim_{h\rightarrow 0}\frac{\sqrt{(x+h)^{2}+1}-\sqrt{x^{2}+1}}{h}\cdot \frac{\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1}}{\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1}} \\ & = \lim_{h\rightarrow 0} \frac{(x+h)^{2}+1-(x^{2}+1)}{h(\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1}} = \lim_{h\rightarrow 0} \frac{x^{2}+2hx+h^{2}+1-x^{2}-1}{h(\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1}}\\ & = \lim_{h\rightarrow 0} \frac{2hx+h^{2}}{h(\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1}} = \lim_{h\rightarrow 0} \frac{2x+h}{(\sqrt{(x+h)^{2}+1}+\sqrt{x^{2}+1}}\\ & =\frac{2x+0}{(\sqrt{(x+0)^{2}+1}+\sqrt{x^{2}+1}}=\frac{2x}{2\sqrt{x^{2}+1}}=\frac{x}{\sqrt{x^{2}+1}}. \end{align*}
  4. Let \[f(x)=\left\{ \begin{array}{cc} \frac{\sin x}{x}-e^{x} & x> 0 \\ x+a & x\leq 0\end{array}\right.\] Solve for $a$ so that $f$ is continuous at $0$.
    Answer: Recall that $f(x)$ is continuous at $0$ if $\lim_{x\rightarrow 0} f(x)$ exists and equals $f(0)$, which in this case is $0+a=a$. For the limit to exists, its left and right limits must agree. If we take the limit as $x$ approaches $0$ from the right, then $f(x)$ will equal $\frac{\sin x}{x}-e^{x}$, and so \[\lim_{x\rightarrow 0^{+}}f(x)=\lim_{x\rightarrow 0^{+}}\left(\frac{\sin x}{x} -e^{x}\right)=1-1=0,\] recalling that $\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$. From the left, \[\lim_{x\rightarrow 0^{-}}f(x)=\lim_{x\rightarrow 0^{-}}(x+a)=0+a=a.\] Thus, for this function to be continuous at zero, the left and right limits must agree and equal zero, and this only works if $a=0$.
  5. Compute the derivatives of the following functions using whatever rules you like.
    1. $\sqrt{1+\sqrt{1+\cos x}}$
      Answer: We use chain rule twice, \begin{align*} (\sqrt{1+\sqrt{1+\cos x}})' & =\frac{1}{2\sqrt{1+\sqrt{1+\cos x}}}\cdot (1+\sqrt{1+\cos x})' =\frac{1}{2\sqrt{1+\sqrt{1+\cos x}}}\left(0+\frac{1}{2\sqrt{1+\cos x}}(1+\cos x)'\right) \\ & =\frac{1}{2\sqrt{1+\sqrt{1+\cos x}}}\frac{1}{2\sqrt{1+\cos x}}(-\sin x). \end{align*}
    2. $e^{x^{2}}\ln \sin (x^{2}+1)$
      Answer: We first use product rule, \[(e^{x^{2}}\ln \sin (x^{2}+1))'=(e^{x^{2}})'\ln \sin (x^{2}+1)+e^{x^{2}}(\ln\sin (x^{2}+1))'.\] For the fist derivative, \[(e^{x^{2}})'=2xe^{x^{2}}\] by the chain rule. For the second derivative, we again use the chain rule twice: (\ln \sin (x^{2}+1))'=\frac{1}{\sin (x^{2}+1)}(\sin (x^{2}+1))'=\frac{1}{\sin(x^{2}+1)}(\cos(x^{2}+1))2x=\frac{2x\cos(x^{2}+1)}{\sin(x^{2}+1)},\] and putting it all together, we get that the main derivative is \[2xe^{x^{2}}\ln \sin (x^{2}+1)+e^{x^{2}}\frac{2x\cos(x^{2}+1)}{\sin(x^{2}+1)} \]
  6. Use implicit differentiation to solve for $\frac{ dy}{dx}$ where $y$ and $x$ are related by the equation \[\frac{x}{y}=\sin \left(\frac{\pi xy}{2}\right)\] when $(x,y)=(1,1)$.
    Answer: This was 7c from Homework 4, the solution is posted there.
  7. Compute the derivatives of the following functions using whatever rules you like
    1. $x^{x}(1+x)^{x^{2}}$
      Answer: Denote this function by $f(x)$. Let us first differentiate $\ln f(x)$, since this function can be simplified as follows: \[\ln f(x)= \ln(x^{x}(1+x)^{x^{2}})=\ln x^{x}+\ln(1+x)^{x^{2}}=x\ln x+x^{2}\ln(1+x)\] Differentiating both sides, we get \[(\ln f(x))'=\frac{f'(x)}{f(x)}=(\ln x+x\frac{1}{x})+(2x\ln x+x^{2}\frac{1}{1+x})=(1+2x)\ln x + 1 +\frac{x^{2}}{1+x}\] Hence \[f'(x)=f(x)(\ln f(x))'=(x^{x}(1+x)^{x^{2}})((1+2x)\ln x + 1 +\frac{x^{2}}{1+x}).\]
    2. $(\cos x + \sin x)^{\cos x + \sin x}$
      Answer: Note that \[(\cos x+\sin x)^{\cos x+ \sin x} = f(g(x))\] where $f(x)=x^{x}$ and $g(x)=\cos x+\sin x$. Hence, we are going to use chain rule, so let us compute the derivatives of $f$ and $g$ first: \[f'(x)=(x^{x})'=(e^{\ln x^{x}})'=e^{\ln x^{x}}(\ln x^{x})'=x^{x}(x\ln x)'=x^{x}(\ln x+1)\] and \[g'(x)=-\sin x+\cos x,\] thus \[(f(g(x)))'=f'(g(x))g'(x)=(\cos x+\sin x)^{\cos x+\sin x}(\ln(\cos x+\sin x)+1)(-\sin x+\cos x).\]
  8. Compute the following limits:
    1. $\lim\limits_{x\rightarrow 0} (\ln x)^{1/(\ln x)}$
      Answer: This is an indeterminate limit. To handle this, we compute the limit of $\ln$ of this, which is \[\ln(\ln x)^{\frac{1}{\ln x}}=\frac{1}{\ln x}\ln\ln x=\frac{\ln\ln x}{\ln x}.\] If we compute the limit of this as $x\rightarrow 0$, the top and bottom go to infinity, so we can use l'Hospital's rule to get \[\lim_{x\rightarrow 0} \frac{\ln \ln x}{\ln x}=\lim_{x\rightarrow 0} \frac{\frac{1}{\ln x}\cdot\frac{1}{x}}{\frac{1}{x}}=\lim_{x\rightarrow 0}\frac{1}{\ln x},\] and since $\lim_{x\rightarrow 0}\ln x=-\infty$, the above limit is zero. Hence \[\lim (\ln x)^{\frac{1}{\ln x}}=\lim_{x\rightarrow 0} e^{\ln (\ln x)^{\frac{1}{\ln x}}}=e^{\lim_{x\rightarrow 0} \ln(\ln x)^{\frac{1}{\ln x}}}=e^{0}=1.\]
    2. $\lim\limits_{x\rightarrow\infty} \frac{(\ln x)^{2}}{x^{2}}$.
      Answer: This limit is indeterminate, so we again use l'Hospital's rule, so taking derivatives of the top and bottom give \[\lim_{x\rightarrow\infty}\frac{(\ln x)^{2}}{x^{2}} =\lim_{x\rightarrow \infty}\frac{\frac{\ln x}{x}}{2x} = \lim_{x\rightarrow \infty}\frac{\ln x}{2x^{2}}.\] Again, this limit is indeterminate, so we again use l'Hospital's rule to differentiate the top and bottom and get that this is equal to \[\lim_{x\rightarrow \infty} \frac{\frac{1}{x}}{4x}=\lim_{x\rightarrow \infty} \frac{1}{4x^{2}}=0.\]
  9. Let \[ f(x)=x^{3}-3x^{2}\]
    1. Find the extreme points of $f$.
      Answer: The extreme points are either where $f'(x)=0$ or where the derivative does not exist. This function is differentiable everywhere, so only the former type of points are possible, hence we sovle \[0=f'(x)=3x^{2}-6x=3x(x-2),\] so the critical points are $0$ and $2$.
    2. Determine which extreme points are local maxima or minima. To determine this, we compute $f''$ at $x=0$ and $x=2$: \[f''(x)=6x-6,\] \[f''(0)=-6<0\] and \[f''(2)=6>0,\] hence $0$ is a local maximum and $2$ is a local minimum.
    3. Find the inflection points.
      Answer: The inflection points are where the second derivative is zero, so we solve \[0=f''(x)=6x-6=6(x-1),\] and this is zero only when $x=1$, so $1$ is an inflection point.
  10. Sketch the graph of the function \[f(x)=\frac{x}{x+1}\] You can use the facts that \[f'(x)=\frac{1}{(x+1)^{2}}\] and \[f''(x)=-\frac{2}{(x+1)^{3}}.\] Graph this on a calculator to see if you were close to correct (however, I will grade on how you determined how to graph this...if you didn't show any work, then you receive no credit).