MATH101 Calculus Workshop Exercise 3 Solutions

W3.1. By the quotient rule, we can differentiate $\cot x$ whenever $\sin x\neq 0$ and obtain

$$\eqalign{{{d}\over{dx}}\cot x&={{d}\over{dx}}{{\cos x}\over{\sin x}}\cr&={{({{d}\over{dx}}\cos x)\sin x-\cos x({{d}\over{dx}}\sin x)}\over{\sin^{2}x}}\cr&={{-\sin^{2}x-\cos^{2}x}\over{\sin^{2}x}}\cr&={{-1}\over{\sin^{2}x}}=-{\hbox{cosec}}^{2}x\qquad(x\neq n\pi)\cr}$$

W3.2 (i) $f(x)=x^{1/2}-6x^{-3}+\log x+\log 2$, so $f^{\prime}(x)=2^{-1}x^{-1/2}+18x^{-4}+x^{-1}.$

(ii) We have ${{d}\over{du}}e^{u}=e^{u}$ and ${{d}\over{dx}}3x^{5}=15x^{4}$, so by the chain rule

$${{d}\over{dx}}e^{3x^{5}}=15x^{4}e^{3x^{5}}.$$

(iii) By the product rule, and then the chain rule, we have

$$\eqalign{{{d}\over{dx}}\Bigl(e^{x^{2}}\sin x\Bigr)&=\Bigl({{d}\over{dx}}e^{x^{2}}\Bigr)\sin x+e^{x^{2}}{{d}\over{dx}}\sin x\cr&=e^{x^{2}}\bigl(2x\sin x+\cos x\bigr).\cr}$$

(iv) By the chain rule

$${{d}\over{dx}}\bigl(\cos 3x\bigr)^{5}=15\cos^{4}3x(-\sin 3x)=-15\sin 3x\cos^{4}3x.$$

(v) By the quotient rule,

$$\eqalign{{{d}\over{dx}}{{\cos x^{4}}\over{5x}}&={{({{d}\over{dx}}\cos x^{4})x-\cos x^{4}{{dx}\over{dx}}}\over{5x^{2}}}\cr&={{-4x^{4}\sin x^{4}-\cos x^{4}}\over{5x^{2}}}.\cr}$$

W3.3 Let $N$ be the number of bacteria, and regard $N$ as a continuous variable depending upon time $x$. The differential equation is

$${{dN}\over{dx}}=kN$$

with initial condition $N(0)=A$, where $k$ and $A$ are constants to be determined. The solution is $N(x)=Ae^{kx}$. Now $N(0)=A=10^{6}$, and $N(100)=2\times 10^{6}$, so $2=e^{100k}$ and $k=(\log 2)/100.$ Now we look for $x$ such that $N(x)=10^{7}$, so $e^{kx}=10$; so

$$x=(\log 10)/k=100(\log 10)/(\log 2)=332s.$$

W3.4 The inverse function rule asserts that ${{dy}\over{dx}}=1\big/{{dx}\over{dy}}.$

(i) When $y=\sin^{-1}x$ we have $x=\sin y$, so ${{dx}\over{dy}}=\cos y$ and $\cos^{2}y+\sin^{2}y=1$. Hence, for $-1<x<1$ we have

$${{dy}\over{dx}}=1\big/\,\,{{dx}\over{dy}}={{1}\over{\cos y}}={{1}\over{\sqrt{1-\sin^{2}y}}}={{1}\over{\sqrt{1-x^{2}}}}.$$

(ii) When $y=\cosh^{-1}x$ we have $x=\cosh y$, so ${{dx}\over{dy}}=\sinh y,$ where $\cosh^{2}y-\sinh^{2}y=1$. Hence for $x>1$ we have

$${{dy}\over{dx}}=1\big/\,\,{{dx}\over{dy}}={{1}\over{\sinh y}}={{1}\over{\sqrt{\cosh^{2}y-1}}}={{1}\over{\sqrt{x^{2}-1}}}.$$

(iii) We take the positive square root in (i) and (ii) since $\sin^{-1}$ and $\cosh^{-1}$ are increasing functions.

consider this point in detail with reference to the graphs

W3.5. The derivative of $\tanh y$ is ${\hbox{sech}}^{2}y$ since by the quotient rule

$$\eqalign{{{d}\over{dy}}\tanh y&={{d}\over{dy}}{{\sinh y}\over{\cosh y}}\cr&={{({{d}\over{dy}}\sinh y)\cosh y-\sinh y({{d}\over{dy}}\cosh y)}\over{\cosh^{2}y}}\cr&={{\cosh^{2}y-\sinh^{2}y}\over{\cosh^{2}}}\cr&={{1}\over{\cosh^{2}y}}.\cr}$$

[1] mark

Let $y=\tanh^{-1}x$ so that $x=\tanh y$ and then by the inverse function rule

$${{dy}\over{dx}}=1\big/\,\,{{dx}\over{dy}}=1\big/\,{\hbox{sech}}^{2}y.$$

Dividing the relation $\cosh^{2}y-\sinh^{2}y=1$ through by $\cosh^{2}y$, we deduce that $1-\tanh^{2}y={\hbox{sech}}^{2}y$, whence

$${{dy}\over{dx}}={{1}\over{1-\tanh^{2}y}}={{1}\over{1-x^{2}}}\qquad(-1<x<1).$$

W3.6. The volume of the cube is $V=x^{3}$, so by the chain rule

$$k={{dV}\over{dt}}={{dV}\over{dx}}{{dx}\over{dt}}=3x^{2}{{dx}\over{dt}}.$$

The surface area is $A=6x^{2}$, so we rearrange this to

$${{dx}\over{dt}}={{k}\over{3x^{2}}}={{2k}\over{A}}.$$

Remark. We expect $A$ to be on the denominator, and $k$ on the numerator; whereas the factor $2$ is a special feature of the cube.

W3.7 (i) Now $f(x)=e^{x}$ has $f^{\prime}(x)=e^{x}$ and $f^{\prime\prime}(x)=e^{x}>0$, so $f$ is convex.

(ii) Also $g(x)=x^{k}$ has $g^{\prime}(x)=kx^{k-1}$ and $g^{\prime\prime}(x)=k(k-1)x^{k-2}>0$, so $g$ is convex for $x>0$.

(iii) Finally, $h(x)=x\log x$ has $h^{\prime}(x)=\log x+1$ and $h^{\prime\prime}(x)=1/x>0$ for $x>0$, so $h$ is convex.

W3.8 Let $P_{n}$ be the statement

$$P_{n}:{{d^{n}}\over{dx^{n}}}(1-x)^{-1}={{n!}\over{(1-x)^{n+1}}}\qquad(-1<x<1).$$

Basis of induction:$P_{1}$ asserts that

$${{d}\over{dx}}(1-x)^{-1}={{1!}\over{(1-x)^{2}}}\qquad(-1<x<1),$$

which is true.

Induction step: Assume $P_{n}$ so that

$${{d^{n}}\over{dx^{n}}}(1-x)^{-1}={{n!}\over{(1-x)^{n+1}}}\qquad(-1<x<1),$$

then differentiate both sides with respect to $x$

$${{d}\over{dx}}{{d^{n}}\over{dx^{n}}}(1-x)^{-1}={{(n+1)n!}\over{(1-x)^{n+2}}}\qquad(-1<x<1),$$

so

$${{d^{n+1}}\over{dx^{n+1}}}(1-x)^{-1}={{(n+1)!}\over{(1-x)^{n+2}}}\qquad(-1<x<1),$$

hence $P_{n+1}$ holds; hence result by induction.

W3.9 (i) The function $\cosh x$ is strictly increasing on $(0,\infty)$, so so ${\hbox{sech}}x=1/\cosh x$ is strictly decreasing. Alternatively, we note that

$${{d}\over{dx}}{\hbox{sech}}x={{d}\over{dx}}{{1}\over{\cosh x}}=-{{\sinh x}\over{\cosh^{2}x}}<0$$

for all $x>0$.

(ii) Dividing the identity $\cosh^{2}y-\sinh^{2}y=1$ by $\cosh^{2}y$, we obtain $1-\tanh^{2}y={\hbox{sech}}^{2}y$. Write $y={\hbox{sech}}^{-1}x$, so $x={\hbox{sech}}y$. Then

$${{dx}\over{dy}}=-\tanh y\,{\hbox{sech}}y=-x\sqrt{1-x^{2}},$$

so by the inverse function rule

$${{dy}\over{dx}}=1/{{dx}\over{dy}}={{-1}\over{x\sqrt{1-x^{2}}}}.$$

W3.10 With

$$y(x)={{1}\over{a}}\cosh ax$$

we have

$${{dy}\over{dx}}=\sinh ax$$

and

$${{d^{2}y}\over{dx^{2}}}=a\cosh ax$$

so

$${{d^{2}y}\over{dx^{2}}}=a\sqrt{1+\sinh^{2}ax}$$

which gives the required result

$${{d^{2}y}\over{dx^{2}}}=a\sqrt{1+\Bigl({{dy}\over{dx}}\Bigr)^{2}}.$$

W3.11 With $g(x)=kx+1+x-e^{x}$ we have $g(x)\rightarrow-\infty$ as $x\rightarrow\infty$ and $g(x)\rightarrow-\infty$ as $x\rightarrow-\infty$. So the maximum value of $g$ occurs at a stationary point. We have

$$g^{\prime}(x)=k+1-e^{x},$$

and

$$g^{\prime\prime}(x)=-e^{x};$$

so there is a stationary point where $g^{\prime}(a)=0$, so $e^{a}=k+1$ and $a=\log(1+k);$ this is a local maximum since $g^{\prime\prime}(a)<0$. Now the maximum value of $g$ is

$$\eqalignno{g(a)&=ka+1+a-e^{a}\cr&=ka+1+a-(k+1)\cr&=(k+1)(a-1)+1\cr&=(k+1)(\log(1+k)-1)+1.\cr}$$

W3.12. (i) The difference quotient is

$$\eqalignno{{{f(a+h)-f(a)}\over{h}}&={{2(a+h)^{2}+5(a+h)-2a^{2}-5a}\over{h}}\cr&={{2a^{2}+4ah+2h^{2}+5a+5h-2a^{2}-5a}\over{h}}\cr&={{4ah+2h^{2}+5h}\over{h}}\cr&=4a+2h+5\cr&\rightarrow 4a+5\qquad(h\rightarrow 0)\cr}$$

so $f$ is differentiable and $f^{\prime}(a)=4a+5.$

(ii) Here the difference quotient is only defined when $a\neq-2$, and for $a\neq-2$ we can choose $a+h\neq-2$ also. Hence taking terms over a common denominator, we have

$$\eqalignno{{{g(a+h)-g(a)}\over{h}}&={{{{1}\over{2+a+h}}-{{1}\over{2+a}}}\over{h}}\cr&={{2+a-(2+a+h)}\over{h(2+a)(2+a+h)}}\cr&=-{{1}\over{(2+a)(2+a+h)}}\cr&\rightarrow{{-1}\over{(2+a)^{2}}}\qquad(h\rightarrow 0)\cr}$$

hence $g$ is differentiable, and $g^{\prime}(a)=-1/(2+a)^{2}.$

W3.13 To find the equation of the tangent to the curve $y=\sin x+\cos x+1$ at $x=3\pi/4$, we note that $y(3\pi/4)=1$ and

$$y^{\prime}=\cos x-\sin x$$

so $y^{\prime}(3\pi/4)=-\sqrt{2}$; hence the tangent is

$$y-1=-\sqrt{2}(x-3\pi/4).$$

W3.14 (i) The function $\sinh x$ is strictly increasing on $(0,\infty)$, so so ${\hbox{cosech}}x=1/\sinh x$ is strictly decreasing. Alternatively, we note that

$${{d}\over{dx}}{\hbox{cosech}}x={{d}\over{dx}}{{1}\over{\sinh x}}=-{{\cosh x}\over{\sinh^{2}x}}<0$$

for all $x>0$.

(ii) From the identity $\cosh^{2}y-\sinh^{2}y=1$, we obtain $\coth^{2}y-1={\hbox{cosech}}^{2}y$. Write $y={\hbox{cosech}}^{-1}x$, so $x={\hbox{cosech}}y$. Then

$${{dx}\over{dy}}=-\coth y\,{\hbox{cosech}}y=-x\sqrt{x^{2}-1},$$

so by the inverse function rule

$${{dy}\over{dx}}=1/{{dx}\over{dy}}={{-1}\over{x\sqrt{x^{2}-1}}}.$$