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\def\dsp{\displaystyle}


\title{Differentiation}

\begin{document}

\maketitle
\begin{enumerate}



\item[Solutions:]

\begin{flushleft}

%%1%%
\item [1.]f is differentiable at a
\item [] $\it \Rightarrow f'(a)=\dsp \lim_{x\rightarrow a} \frac{\displaystyle f(x) - f(a)}{\displaystyle x -
a}$   exists
\item [] $\it \Rightarrow \dsp \lim_{x\rightarrow a}(f(x)-f(a)) = \dsp \lim_{x\rightarrow a} \frac{\displaystyle f(x) - f(a)}{\displaystyle x -
a}(x-a)$
\item [] \hspace{8.65em} $=\dsp \lim_{x\rightarrow a} \frac{\displaystyle f(x) - f(a)}{\displaystyle x -
a}\dsp \lim_{x\rightarrow a}(x-a)$
\item [] \hspace{8.65em} = $\it f'(a)\cdot 0$
\item [] \hspace{8.65em} = 0
\item [] $\it \Rightarrow \dsp \lim_{x\rightarrow a}f(x)=f(a)$
\item [] $\Rightarrow$ f is continuous at a.

\

%%2%%
\item [2.] (For example)\quad \enspace $\it f(x)=\mid x\mid$
\item [] Then f is continuous at 0 , but f is not differentiable at
0

\

%%3%%
\item [3.]f is differentiable at 1   \\
 $\Rightarrow$    f is continuous at 1 \\
$\Rightarrow  \dsp \lim_{x\rightarrow 1^{-}}ax^2+bx+7=\dsp
\lim_{x\rightarrow 1^{+}}2bx+2a $ \\
$\Rightarrow a+x+7=2b+2a$ \\
$\Rightarrow a + b = 7 \hspace{6em} \cdots(1) $

\

\item[]\hspace{0.5em}     f is differentiable at 1 \\

\

$\Rightarrow \it \left\{ \begin{array}{ll}
           f'(1)=\frac{\displaystyle d}{\displaystyle
           dx}(ax^2+bx+7) \mid_{x=1} \\
\hspace{2.4em} =2ax+b \mid_{x=1} \\
\hspace{2.4em} =2a+b \\
           f'(1)=\frac{\displaystyle d}{\displaystyle
           dx}(2bx+2a) \mid_{x=1} \\
\hspace{2.4em} =2b \mid_{x=1} \\
\hspace{2.4em} =2b \\
        \end{array} \right.$ \\

      \bigskip

$\Rightarrow    2a +b = 2b $\\
$\Rightarrow      2a- b = 0  \hspace{5.5em} \cdots(2)$ \\
\medskip
(1),(2) $\Rightarrow  \left\{ \begin{array}{ll}
           a + b = 7 \\
           2a - b = 0
        \end{array} \right.
$ \\
\medskip
$\Rightarrow  a=\frac{\displaystyle 7}{\displaystyle 3}
\enspace,\enspace b=\frac{\displaystyle 14}{\displaystyle 3} $

$$ $$

%%4%%
\item [4.] $(1) \frac{\displaystyle dy}{\displaystyle
dx}=3\sin^2(\tan x)\cos(\tan x)\sec^2x $\\
%%
$ \hspace{0.75em} (2)\frac{\displaystyle dy}{\displaystyle
dx}=\sec(\cos^{\frac{1}{ 2}}(x^2+5-\frac{\displaystyle
1}{\displaystyle x}))\tan(\cos^{\frac{1}{
2}}(x^2+5-\frac{\displaystyle 1}{\displaystyle x})) \cdot $\\
$\hspace{4.5em} \frac{\displaystyle 1}{\displaystyle
2}\cos^{-\frac{1}{ 2}}(x^2+5-\frac{\displaystyle 1}{\displaystyle
x}) \cdot (-\sin(x^2+5-\frac{\displaystyle 1}{\displaystyle x}))
\cdot
(2x+x^{-2}) $ \\
%%
\hspace{0.75em}(3)$\frac{\displaystyle dy}{\displaystyle
dx}=2x\cot(\csc x) + x^2(-\csc^2(\csc x)(-\csc x\cot x))$ \\
$\hspace{4.5em}+ \sin x\cos
x +x\cos x\cos x+x\sin x(-\sin x)$ \\
\hspace{3.3em} $=2x\cot(\csc x) + x^2\csc^2 (\csc x)\csc x \cot x
$\\
\hspace{4.15em} $+ \sin x\cos x + x\cos^2 x - x\sin^2 x$ \\
%%
\hspace{0.75em}(4)$\frac{\displaystyle dy}{\displaystyle
dx}=(\cos(x^2+1))\cdot 2x  + \frac{\displaystyle (x^3+1)-(x+2)\cdot
3x^2}{\displaystyle (x^3+1)^2} $ \\
\hspace{3.3em}  $=2x\cos(x^2+1)+\frac{\displaystyle
-2x^3-6x^2+1}{\displaystyle (x^3+1)^2} $

\

%%5%%
\item [5.] $ y'=n{a_{n}}x^{n-1} + (n-1)a_{n-1}x^{n-2}+\cdots+3{a_3}x^2+ 2a_2x +a_1$\\
$\hspace{0.75em}y''=n(n-1) a_nx^{n-2} + (n-1)(n-2) a_{n-1} x^{x-3} +
\cdots +
3\cdot2 a_3x + 2a_2 $ \\
$\hspace{0.75em}y^{(n)}=a_n\cdot n! $ \\
$\hspace{0.75em}y^{(n+1)}=0 $

\

%%6%%
\item [6.] $y^3=x^2+\cos \frac{\displaystyle x}{\displaystyle y} $\\
$\Rightarrow \frac{\displaystyle d}{\displaystyle dx}(y^3)=
\frac{\displaystyle d}{\displaystyle dx}(x^2+\cos
\frac{\displaystyle x}{\displaystyle y})$ \\
$\Rightarrow 3y^2 \frac{\displaystyle dy}{\displaystyle dx}=2x -
(\sin \frac{\displaystyle x}{\displaystyle y})\cdot
\frac{\displaystyle y-x \frac{\displaystyle dy}{\displaystyle
dx}}{\displaystyle y^2}$ \\
\hspace{4em} $=2x - (\sin \frac{\displaystyle x}{\displaystyle
y})\cdot \frac{\displaystyle 1}{\displaystyle y} + (sin
\frac{\displaystyle x}{\displaystyle y})\cdot \frac{\displaystyle
x}{\displaystyle y^2}\cdot \frac{\displaystyle dy}{\displaystyle dx}
$ \\
$\Rightarrow (3y^2- \frac{\displaystyle x}{\displaystyle y^2}\sin
\frac{\displaystyle x}{\displaystyle y}) \frac{\displaystyle
dy}{\displaystyle dx}=2x- \frac{\displaystyle 1}{\displaystyle
y}\sin \frac{\displaystyle x}{\displaystyle y}$ \\
$\Rightarrow \frac{\displaystyle dy}{\displaystyle dx} =
\frac{\displaystyle 2x - \frac{\displaystyle 1}{\displaystyle y}\sin
\frac{\displaystyle x}{\displaystyle y}}{\displaystyle 2y^2 -
\frac{\displaystyle x}{\displaystyle y^2}\sin \frac{\displaystyle
x}{\displaystyle y}} $

\

\


%%7%%
\item [7.] $xy =\sin x+\cos y $\\
$\Rightarrow \displaystyle \frac{d}{dx}(xy)=\frac{d}{dx}(\sin x
+\cos y) $ \\
$\Rightarrow \displaystyle y + x\frac{dy}{dx}=\cos x-\sin y\frac{dy}{dx} $\\
$\Rightarrow \displaystyle (x+\sin y)\frac{dy}{dx}=\cos x - y $ \\
${}_\cdot{}^\cdot{}_\cdot \enspace \displaystyle
\frac{dy}{dx}=\frac{\cos x-y}{x+\sin y}$ \\
$\Rightarrow \displaystyle
\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dx}(\frac{\cos
x-y}{x+\sin y}) $ \\
$=\displaystyle \frac{(-\sin x - \frac{\displaystyle
dy}{\displaystyle dx})(x + \sin y ) - ( 1 +
\cos y \frac{\displaystyle dy}{\displaystyle dx} )( \cos x - y )} {(x + \sin y)^2} $ \\
=$\displaystyle \frac{-x\sin x - \sin x \sin y - \cos x + y -
(x+\sin y + \cos y\cos x - y\cos y )\frac{\displaystyle
dy}{\displaystyle dx}}{(x + \sin y)^2} $ \\
=$\displaystyle \frac{-x\sin x - \sin x \sin y - \cos x + y -
(x+\sin y + \cos y\cos x - y\cos y )\frac{\displaystyle
\cos x - y}{\displaystyle x + \sin y}}{(x + \sin y)^2} $ \\

\

\

%%8%%
\item [8.]\item [(1)] $ \left\{ \begin{array}{ll}
           2^4 - 4\cdot2^2 = 0 \\
           3^4 - 9\cdot3^3 = 0
        \end{array} \right.$ \\

        \

$\Rightarrow 2^4-4\cdot2^2=3^4-9\cdot3^2$\\
$\Rightarrow$ (3,2) is on the curve C. \\
Similarly, for (3,-2). \\

\

\item [(2)] $ y^4-4y^2=x^4-9x^2 $\\
$\Rightarrow \frac{\displaystyle d}{\displaystyle dx}(y^4-4y^2) =
\frac{\displaystyle d}{\displaystyle dx}(x^4-9x^2) $ \\
$\Rightarrow 4y^3\frac{\displaystyle dy}{\displaystyle dx} -
8y\frac{\displaystyle dy}{\displaystyle dx} = 4x^3 - 18x $ \\
$\Rightarrow \frac{\displaystyle dy}{\displaystyle dx} =
\frac{\displaystyle 4x^3 - 18 x}{\displaystyle 4y^3 - 8y} $ \\

\

\hspace{2.8em}=$ \left\{ \begin{array}{ll}
           \hspace{0.8em} \frac{\displaystyle 27}{\displaystyle 8} \hspace{3em}(x,y)=(3,2)\\
           -\frac{\displaystyle 27}{\displaystyle 8} \hspace{3em}(x,y)=(3,-2)
        \end{array} \right.$ \\

        \

${}_\cdot{}^\cdot{}_\cdot \enspace Equation: y -2
=\frac{\displaystyle 27}{\displaystyle 8}(x-3)$

\

\item [(3)]slope of the tangent $ \cdot$ slope of the normal = -1 \\
$\Rightarrow -\frac{\displaystyle 27}{\displaystyle 8}\enspace \cdot
\enspace$ slope of the normal = -1 \\
$\Rightarrow $ slope of the normal = $\frac{\displaystyle
8}{\displaystyle 27}$ \\
${}_\cdot{}^\cdot{}_\cdot \enspace Equation: y +2
=\frac{\displaystyle 8}{\displaystyle 27} (x-3) $

\

%%9%%
\item [9.](1) $\frac{\displaystyle dy}{\displaystyle dx}=\frac{\displaystyle \frac{\displaystyle dy}{\displaystyle dt}}{\displaystyle \frac{\displaystyle dy}{\displaystyle
dt}}$ \\
\hspace{4em}= $\frac{\displaystyle \sin t}{\displaystyle 1 - \cos t} $ \\
\hspace{2.1em}$\frac{\displaystyle dy}{\displaystyle dx}
\mid_{t=\frac{\pi}{3}}=\frac{\displaystyle \sin \frac{\displaystyle
\pi}{\displaystyle 3}}{\displaystyle 1 - \cos \frac{\displaystyle
\pi}{\displaystyle 3}} $ \\
\hspace{5.3em}=$\frac{\displaystyle \frac{\displaystyle
\sqrt{3}}{\displaystyle
2}}{\displaystyle 1 - \frac{\displaystyle 1}{\displaystyle 2}}$ \\
\hspace{5.3em}=$\sqrt{3}$

$$ $$

\item []$ \frac{\displaystyle d^2y}{\displaystyle dx^2}= \frac{\displaystyle d(\frac{\displaystyle dy}{\displaystyle dx})}{\displaystyle
dx}$ \\

\

\hspace{2em}=$\frac{\displaystyle dy'}{\displaystyle dx}$ \\

\

\hspace{2em}=$\frac{\displaystyle \frac{\ dy'}{\
dt}}{\displaystyle \frac{\ dx}{ dt}} $ \\

\

\hspace{2em}=$\frac{\displaystyle \frac{\displaystyle \cos t (1 -
\cos t ) - \sin t \cdot \sin t}{\displaystyle (1 - \cos t
)^2}}{\displaystyle 1
- \cos t} $ \\

\

\hspace{2em}=$\frac{\displaystyle \cos t - 1}{\displaystyle (1 -
\cos t
)^2}\cdot\frac{\displaystyle 1}{\displaystyle 1 - \cos t} $ \\

\

\hspace{2em}= - $\frac{\displaystyle 1}{\displaystyle (1- \cos t )^2} $ \\

$$ $$

$\frac{\displaystyle d^2y}{\displaystyle dx^2}
\mid_{t=\frac{\pi}{3}}=-\frac{\displaystyle 1}{\displaystyle (1 -
\cos t )^2} $ \\

\

\hspace{3.35em} =$-\frac{\displaystyle 1}{\displaystyle 4} $

$$ $$
\item [](2) t$=\frac{\displaystyle
\pi}{\displaystyle 3}$ \\

\

\hspace{1.6em}$x=\frac{\displaystyle \pi}{\displaystyle 3}-\sin
\frac{\displaystyle \pi}{\displaystyle 3} = \frac{\displaystyle
\pi}{\displaystyle 3} - \frac{\displaystyle \sqrt{3}}{\displaystyle
2} $ \\

\

\hspace{1.6em}$y=1 - \cos \frac{\displaystyle \pi}{\displaystyle 3}
=
\frac{\displaystyle 1}{\displaystyle 2} $ \\

\

\hspace{1.6em}Equation :  $y -\frac{\displaystyle 1}{\displaystyle
2} = \sqrt{3}(x - \frac{\displaystyle \pi}{\displaystyle 3} +
\frac{\displaystyle \sqrt{3}}{\displaystyle 2}) $

$$ $$

%%10%%
\item [10.] Let x = distance between A and the intersection \\
         \hspace{2.8em}       y = distance between B and the intersection \\
         \hspace{2.9em}       s = distance between A and B \\

\
  \begin{figure}[h]


 \quad\includegraphics[scale=0.75]{fig2.eps}\\~\\



\end{figure}



%%%%%%%%%šĪ%%%%%%%%%

Then $ \left\{ \begin{array}{ll}
           \tan \theta = \displaystyle \frac{x}{y}   \\
           s^2 = x^2 + y^2
        \end{array} \right.$ \\

        \

Now $\displaystyle \frac{dx}{dt}$ = - 3 m / $\sec$  , $\displaystyle
\frac{dy}{dt} $ = 2 m / $\sec $ \\

\

\hspace{2.2em}

\


  $\tan \theta = \displaystyle
\frac{x}{y}$ \\

$\Rightarrow \displaystyle \frac{d}{dt}\tan \theta = \displaystyle
\frac{d}{dt}\displaystyle (\frac{x}{y}) $ \\

\

$\Rightarrow \sec^2\theta\displaystyle
\frac{d\theta}{dt}=\displaystyle \frac{\displaystyle
\frac{dx}{dt}\cdot y - x \displaystyle \frac{dy}{dt}}{y^2}$ \\

\

\hspace{2.2em}

\


 x = 10 m , y = 25 m \\

\

$\Rightarrow \tan \theta = \displaystyle \frac{10}{25}=\displaystyle
\frac{2}{5}$ \\

\

$\Rightarrow \sec^2\theta = \displaystyle \frac{29}{25} $ \\


\

\hspace{2.2em}

\


${}_\cdot{}^\cdot{}_\cdot \enspace \displaystyle \frac{29}{25}
\displaystyle \frac{d\theta}{dt} = \displaystyle \frac{-3\cdot25 -
10 \cdot 2}{25^2} $ \\

\

 \hspace{4em} $\displaystyle \frac{d\theta}{dt}$= - $\displaystyle \frac{19}{145}$\quad( rad / $\sec$ ) \hspace{2em}
$\cdots (1) $ \\

\

\hspace{2.2em}

\



 $\hspace{1.3em}s^2 = x^2 + y^2$ \\

 \

$\Rightarrow \displaystyle \frac{d}{dt}s^2=\displaystyle
\frac{d}{dt}(x^2 + y^2) $ \\

\

$\Rightarrow 2s\displaystyle \frac{ds}{dt}=2x\displaystyle
\frac{dx}{dt} + 2y\displaystyle \frac{dy}{dt} $ \\

\

$\hspace{1.3em}$x = 10 m , y = 25 m \\

\

$\Rightarrow s =\ \sqrt{10^2+25^2}$ m \\

\hspace{2.1em}= 5 $\sqrt{29}$ m \\

\

${}_\cdot{}^\cdot{}_\cdot \enspace
2\cdot5\sqrt{29}\cdot\displaystyle
\frac{ds}{dt}=2\cdot10\cdot(-3)+2\cdot25\cdot2 $\\

\

\hspace{5.6em}$\displaystyle \frac{ds}{dt}=\displaystyle
\frac{4}{\sqrt{29}}\hspace{1em}$( m / $\sec$ )$\hspace{2em}\cdots(2)
$

$$ $$

%%11%%
\item [11.](1) $f(x)=x^{\frac{1}{3}}$ \\

\

\hspace{1.5em}$\Rightarrow f'(x)= \displaystyle \frac{1}{3}
x^{-
\frac{2}{3}}$ \\

\

\hspace{1.5em}$\Rightarrow  df = f'(x)dx $\\

\

\hspace{4.1em}=$\displaystyle \frac{1}{3} x^{- \frac{2}{3}}dx $ \\

\

\item [](2) Let $f(x)=\sqrt[3]{x}$ \\

\

\hspace{1.5em}Note $f(x+\Delta x)\approx f(x) + f'(x)\Delta x$ \quad
($\Delta$ x
is small)\\

\

\hspace{1.5em}Take x = 1000 , $\Delta x$ = 0.2 \\

\

$f(1000.2) \approx f(1000) + f'(1000)\cdot0.2$ \\

\

\hspace{4.5em}$= 10 +  \displaystyle \frac{1}{3}\cdot \displaystyle
\frac{1}{100}\cdot0.2 $\\

\

\hspace{4.5em}$\approx 10.0006 $\\

\

${}_\cdot{}^\cdot{}_\cdot \enspace \sqrt[3]{1000.2}\approx10.0006$














\end{flushleft}


\end{enumerate}



\end{document}