Midterm 2: problems and solutions

  1. (10 points) Compute the following derivatives:
    1. $\ln ((x+1)^{\frac{1}{3}}\cos^{5}x)$
      Answer: The first thing to do is to use the laws for logarithms to split this up, e.g. $\ln ab+\ln a+\ln b$, $\ln a^{b}=b\ln a$, so that \[\ln((x+1)^{\frac{1}{3}}\cos^{5}x)=\ln (x+1)^{\frac{1}{3}}+\ln\cos^{5}x=\frac{1}{3}\ln(x+1)+5\ln \cos x,\] and now when we differentiate, we differentiate each piece: \[(\frac{1}{3}\ln (x+1))'=\frac{1}{3}(\ln (x+1))'=\frac{1}{3}\frac{1}{x+1}\] and by chain rule, \[(5\ln \cos x)'=5(\ln \cos x)'=5\frac{1}{\cos x}(\cos x)'=5\frac{1}{\cos x}(-\sin x)=-5\frac{\sin x}{\cos x}=-5\tan x,\] and so the final solution is \[\frac{1}{3}\frac{1}{x+1}-5\tan x.\]
    2. $x^{\frac{1}{\ln x}}+e^{\sin x^{2}}$
      Answer: We will differentiate each term separately and add them together in the end. To differentiate the first term, set $f(x)=x^{\frac{1}{\ln x}}$, so that \[\ln f(x)=\ln x^{\frac{1}{\ln x}}=\frac{1}{\ln x}\ln x=1,\] but then \[(\ln f(x))'=\frac{f'(x)}{f(x)}=(1)'=0,\] so $(x^{\frac{1}{\ln x}})'=0$. We could have also seen this by noting that $x^{\frac{1}{\ln x}}=(e^{\ln x})^{\frac{1}{\ln x}}=e^{1}$, which is a constant and hence has derivative zero. For the second term, we use chain rule twice: \[(e^{\sin x^{2}})'=e^{\sin x^{2}}(\sin x^{2})'=e^{\sin x^{2}}(\cos x^{2})(2x),\] and adding these derivatives together, we get that \[(x^{\frac{1}{\ln x}}+e^{\sin x^{2}})'=0+2x\cos x^{2}(e^{\sin x^{2}}).\]
  2. . (10 points) Compute the derivatives of the following functions
    1. $(2x)^{2x}$
      Answer: The easiest way to do this is to use chain rule by observing $(2x)^{2x}=f(g(x))$, where $f(x)=x^{x}$ and $g(x)=2x$. Then $g'(x)=2$, while $f'(x)$ we compute as follows: \[\ln f(x)=\ln x^{x}=x\ln x,\] \[(\ln f(x))'=\frac{f'(x)}{f(x)}= 1\cdot \ln x+ x\frac{1}{x}=\ln x+1 \;\;\; (\mbox{product rule})\] \[f'(x)=f(x)(\ln x+1)=x^{x}(\ln x+1).\] Thus, using the chain rule formula, \[((2x)^{2x})'=(f(g(x)))'=f'(g(x))g'(x)=(g(x))^{g(x)}(\ln g(x)+1)g'(x)=(2x)^{2x}(\ln 2x+1)\cdot 2.\]
    2. $\ln(\sec x+\tan x)$
      Answer: By chain rule, \[(\ln(\sec x+\tan x))'=\frac{1}{\sec x+\tan x}(\sec x+\tan x)'=\frac{1}{\sec x+\tan x}((\sec x)'+(\tan x)'),\] so we just need to compute the derivatives of $\sec x$ and $\tan x$. We use quotient rule for both: \[(\sec x)'=\left(\frac{1}{\cos x}\right)'=\frac{0\cdot \cos x-1\cdot(-\sin x)}{\cos^{2}x}=\frac{\sin x}{\cos^{2}x}=\frac{\sin x}{\cos x}\frac{1}{\cos x}=\tan x \sec x,\] while \[(\tan x)'=\left(\frac{\sin x}{\cos x}\right)'=\frac{(\sin x)'\cos x-\sin x(\cos x)'}{\cos^{2} x}=\frac{\sin^{2} x+\cos^{2} x}{\cos^{2}x}=\frac{1}{\cos^{2} x}=\sec^{2}x,\] and putting it all together, we get \[(\ln(\sec x+\tan x))'=\frac{1}{\sec x+\tan x}(\tan x \sec x+\sec^{2} x)=\frac{1}{\sec x+\tan x}(\tan x+\sec x)\sec x=\sec x.\]
  3. (10 points) Let $f(x)=x^{3}-6x^{2}+17x$. Find all tangent lines that are parallel to $y=2x+3$. (Write them in the usual form $y-y_{0}=m(x-x_{0})$.)
    Answer: Well, if the tangent line at a point $x$ is parallel to $y=2x+3$, that means that they have the same slope, so they both have slope 2. However,t he slope of a tangent line at a point $x$ is just $f'(x)$, so we need to solve for when $f'(x)=2$, that is, \[f'(x)=-3x^{2}-12x+17=2\] \[-3x^{2}-12x+15=0\] \[-3(x^{2}+4x-5)=0\] \[(x+1)(x-5)=0\] In the last line, we divided both sides of the equation by -3 and then factored (you can use the quadratic formula to find the roots). Thus, the tangent lines parallel to $y=2x+3$ are at the points $1$ and $-5$, so now we just need to compute the tangent lines at these points. To compute the tangent line at 1, recall that its formula is \[y-f(1)=f'(1)(x-1).\] Note that we already know what f'(1) is: it's 2. So we just need to compute f(1): \[f(1)=-(1)^{3}-6(1)^{2}+17(1)=10,\] so the first tangent line is \[y-10=2(x-1).\] For the second tangent line, the equation is \[y-f(-5)=2(x-(-5))=2(x+5),\] where \[f(-5)=-(-5)^{3}-6(-5)^{2}+17(-5)=125-150-85=-110,\] and so the second tangent line is \[y+110=2(x+5).\]
  4. (10pts) Suppose you know that $f'(x)=f(x)e^{f(x)}$. Without computing $f^{-1}(x)$, compute $(f^{-1})'(x)$.
    Answer: For this, we just use the formula for the derivative of an inverse, \[(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))},\] and because $f'(x)=f(x)e^{f(x)}$, this above is equal to \[=\frac{1}{f(f^{-1}(x))e^{f(f^{-1}(x))}}=\frac{1}{xe^{x}},\] where in the last equality, we used the fact that $f(f^{-1}(x))=x$.
  5. Suppose a sphere is increasing in size so that when its radius has reached 1, it's surface area is increasing at a rate of $8\pi$ units per second. How fast is its volume increasing at this time? (Recall that the volume of a sphere of radius $r$ is $\frac{4}{3}\pi r^{3}$ and its surface area is $4\pi r^{2}$.)
    Answer: Let $t_{0}$ be the time when the radius is $1$, so if we view the radius $r(t)$ as a function of time, we have $r(t_{0})=1$. If $V(t)=\frac{4}{3}\pi r(t)^{3}$ denotes the volume of the sphere, our objective is to compute how fast it is growing at $t=t_{0}$, i.e. we want to compute $V'(t_{0})$: \[V'(t)=\left(\frac{4}{3}\pi r(t)^{3}\right)'=\frac{4}{3}\pi (3r(t)^{2} r'(t))=4\pi r(t)^{2} r'(t),\] and so \[V'(t_{0})=4\pi r(t_{0})^{2}r'(t_{0})=4\pi (1)^{2} r'(t_{0})=4\pi r'(t_{0}).\] Hence, it remains to figure out what $r'(t_{0})$ is. To do this, we use the other bit of information in our problem, that the surface area is increasing at $8\pi$ units per second at the time $t_{0}$. Letting $A(t)=4\pi r(t)^{2}$ denote the surface area as a function of time, we know that $A'(t_{0})=8 \pi$, where \[A'(t)=(4\pi r(t)^{2})'=4\pi (2 r(t)r'(t))=8 \pi r(t) r'(t),\] and so \[8\pi = A'(t_{0})=8\pi r(t_{0})r'(t_{0})=8\pi(1)r'(t_{0})=8\pi r'(t_{0}).\] Again, we used the fact that $r(t_{0})=1$. Hence $r'(t_{0})=1$, and so \[V'(t_{0})=4\pi r'(t_{0})=4\pi (1)=4\pi.\]