Homework 3, due July 6 at beginning of class.

  1. Compute the derivatives of the following functions using either the power rule, quotient rule, product rule, or chain rule. Below, the symbol $a$ will represent some real number that is independent of $x$. (You do not have to expand your solutions out, keep them as simple as possible).
    1. $x^{2}+5$
    2. $(x^{2}+1)(x^{3}+5)$
      Answer: Apply the quotient rule, the derivative is \[\frac{(2x+0)(x^{3}+5)-(x^{2}+1)(3x^{2}+0)}{(x^{3}+5)^{2}} = \frac{5x^{4}+10x+3x^{2}}{(x^{3}+5)^{2}}.\]
    3. $\frac{1}{x^{1/3}}+\frac{2}{x+a}$
    4. $\frac{1}{\sqrt{x}}$
    5. $x\cos x$
      Answer: Product rule: its derivative is \[1\cdot \cos x+ x\cdot(-\sin x)=\cos x-x\sin x.\]
    6. $\cos ax$
      Answer: By chain rule, \[(\cos ax)'=-a\sin x.\]
    7. $\sin x^{2}$
      Answer: Again, chain rule, \[(\sin x^{2})'=2x \cos x^{2}.\]
    8. $\sin^{2}x$
      Answer: There are two ways to go about this: the first is product rule \[(\sin^{2}x)'=((\sin x)(\sin x))'=\cos x(\sin x)+\sin x(\cos x)=2\sin x\cos x.\] Alternatively, you can also use chain rule: the outside function is $x^{2}$ and the inside function is $\sin x$.
    9. $\cot x$
      Answer: By quotient rule, \[(\cot x)'=\left(\frac{\cos x}{\sin x}\right)'=\frac{(-\sin x)(\sin x)-\cos x(\cos x)}{\sin^{2}x}=\frac{-\sin^{2}x-\cos^{2}x}{\sin^{2}x}=-\frac{1}{\sin^{2}x}.\]
    10. $\sqrt{x^{2}+a^{2}}$
      Answer: Chain rule, the outside function being $f(x)=\sqrt{x}$ with derivative $\frac{1}{2\sqrt{x}}$, and the inside function being $g(x)=x^{2}+a^{2}$ with derivative $2x$, hence, using the chain rule formula, \[(f\circ g)'(x)=f'(g(x))g'(x)=\frac{1}{2\sqrt{x^{2}+a^{2}}}2x=\frac{x}{\sqrt{x^{2}+a^{2}}}.\]
    11. $\frac{x^{2}+5}{x+2}$
    12. $\frac{x}{\sqrt{x^{2}+1}}$
      Answer: This involves quotient rule, but before doing the quotient rule, let's compute the derivative of the denominator, since we will need it. We can use the problem from earlier, letting $a=1$, so \[(\sqrt{x^{2}+1})'=\frac{x}{x^{2}+1}.\] Thus, \[ \left(\frac{x}{\sqrt{x^{2}+1}}\right)' =\frac{1\cdot \sqrt{x^{2}+1}-x\left(\frac{x}{\sqrt{x^{2}+1}}\right)}{\sqrt{x^{2}+1}^{2}} =\frac{\frac{x^{2}+1}{\sqrt{x^{2}+1}}-\frac{x^{2}}{\sqrt{x^{2}+1}}}{x^{2}+1} = \frac{\frac{-1}{\sqrt{x^{2}+1}}}{x^{2}+1}=-\frac{1}{(x^{2}+1)^{\frac{3}{2}}}.\]
    13. $\frac{1}{\cos^{2} x}$.
    14. $\cos(\sin(\cos x))$
    15. $\sqrt{1+\sqrt{1+x^{2}}}$
      Answer: Let $f,g,h$ be functions and let $i(x)=g(h(x))$. By the chain rule, \[(f(g(h(x))))'=(f(i(x)))'=f'(i(x))i'(x)=f'(g(h(x))i'(x)=f'(g(h(x)))g'(h(x))h'(x).\] For this particular problem, our function can be written as $f(g(h(x)))$ where $f(x)=\sqrt{x}$, $g(x)=1+\sqrt{x}$, and $h(x)=1+x^{2}$. $f$ and $g$ both have derivative $\frac{1}{2\sqrt{x}}$, and $h$ has derivative $2x$, hence, using the long formula above that we just came up with, we get that the derivative is \[\frac{1}{2\sqrt{1+\sqrt{1+x^{2}}}}\frac{1}{2\sqrt{1+x^{2}}}2x.\]
    16. $(1+x)^{3}(1+x^{2})^{2}(1+x^{3})$
  2. Find the tangent line to each function at the given point. You may use any rules available to evaluate the derivatives.
    1. $\frac{1}{12}x^{3}$ at $x=2$,
      Answer: Call this function $f(x)$. The tangent line at $x=2$ will be \[y-f(2)=f'(2)(x-2).\] Since $f(2)=\frac{8}{12}=\frac{2}{3}$, and $f'(2)=\frac{3(2)^{2}}{12}=1$, we get that the tangent line is \[y-\frac{2}{3}=x-2.\]
    2. $\frac{5}{x^{4/5}}$ at $x=1$
    3. $\sin x$ at $x=\pi$
    4. $\sqrt{1-x^{2}}$ at $x=\frac{1}{\sqrt{2}}$.
      Answer: Denote this function by $f(x)$. The tangent line for this will be of the form \[y-f'(\frac{1}{\sqrt{2}})=f'(\frac{1}{\sqrt{2}})(x-\frac{1}{\sqrt{2}}).\] Thus, we just have to figure out what $f$ and $f'$ are at $\frac{1}{\sqrt{2}}$. Firstly, \[f(\frac{1}{\sqrt{2}})=\sqrt{1-(\frac{1}{\sqrt{2}})^{2}}=\sqrt{1-\frac{1}{2}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}.\] Moreover, by the chain rule, (with outside function $\sqrt{x}$ and insude function $1-x^{2}$, we get \[f'(x)=-\frac{x}{\sqrt{1-x^{2}}},\] and so \[f'(\frac{1}{\sqrt{2}})=-\frac{\frac{1}{\sqrt{2}}}{\sqrt{1-(\frac{1}{\sqrt{2}})^{2}}} = -\frac{1}{\sqrt{2}}\frac{1}{\sqrt{1-\frac{1}{2}}} =-\frac{1}{\sqrt{2}}\frac{1}{\sqrt{\frac{1}{2}}}=-\frac{1}{\sqrt{2}}\frac{1}{\frac{1}{\sqrt{2}}}=-1,\] hence the tangent line is \[y-\frac{1}{\sqrt{2}}=-x-\frac{1}{\sqrt{2}},\] or alternatively, \[y=-x.\]
  3. Find all tangent lines to the curve $\frac{1}{3}x^{3}-3x^{2}+1$ that have slope equal to $-8$.
    Answer: If $c$ is a point where the tangent line has slope $-8$, remember that the slope of this tangent line is defined to be $f'(c)$, so we need to solve for where $f'(c)=8$. Computing $f'$, we get \[f'(x)=x^{2}-6x\] and we set this equal to $-8$. Adding $8$ to both sides of the resulting equation, we just need to solve \[x^{2}-6x+8=0\] but the roots of this polynomial are $2$ and $4$, hence these are the points where the tangent lines have slope equal to $-8$. Since we know their slopes already, to figure out the tangent lines we need only know what $f(2)$ and $f(4)$ are: \[f(2)=\frac{8}{3}-3\cdot 4+1=\frac{8}{3}-11=-\frac{25}{3},\] and \[f(4)=\frac{64}{3}-3\cdot 16+1=\frac{19}{3},\] and thus the tangent lines at $c=2$ and $c=4$ are, respectively, \[y+\frac{25}{3}=-8(x-2)\] and \[y-\frac{19}{3}=-8(x-4).\]