Continuity and Differentiability

134. Given, $x=a(cost+tsint)$ and $y=a(sint-tcost).$

Differentiating w r t. 't' we get,

$\frac{dx}{dt}=a\frac{d}{dt}(cost+tsint).$

$=a\left(-sint+t\frac{d}{dt}sint+sint\frac{dt}{dt}\right).$

$=a(-sint+tcost+sint)=atcost$

$\frac{dy}{dt}=\frac{ad}{dt}(csint-tcost)$

$=a\left(cost-t\frac{d}{dt}cost-cot\frac{dt}{dt}\right)$

$=a(cost+tsint-cost)=atsint$

$\frac{dy}{dx}=\frac{dy/dt}{dx/at}=\frac{atsint}{atcost}=tant$

So, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(tant)=\frac{d}{dt}(tant)\cdot \frac{dt}{dx}.$

$=se{c}^{2}t*\frac{dt}{dx}.$

$=se{c}^{2}t*\frac{1}{(dx/dt)}$

$=se{c}^{2}t*\frac{1}{ at cost}$

$=\frac{se{c}^{3}t}{at}$

133.  Given, $cosy=xcos(a+y).$

$x=\frac{cosy}{cos(a+y)}$

Differentiating w r t 'y' we get,

$\frac{dx}{dy}=\frac{d}{dy}\left(\frac{cosy}{cos(a+y)}\right).$

$=\frac{cos(a+y)\frac{d}{dy}cosy-cosy\frac{d}{dy}cos(a+y)}{co{s}^2(a+y).}$

$=\frac{cos(a+y)(-siny)-cosy(-sin(a+y))}{co{s}^2(a+y).}$

$=\frac{-cos(a+y)siny+sin(a+y)cosy}{co{s}^2(a+y)}$

$=\frac{sin(a+y)cosy-cos(a+y)siny.}{co{s}^2(a+y)}$

$\frac{dx}{dy}=\frac{sin(a+y-y)}{co{s}^2(a+y)}\left\{\begin{array}{rr}\hfill ?sin(A-B)=sinAcosB& \hfill -cosAsinB\end{array}\right\}$

So, $\frac{dy}{dx}=\frac{co{s}^2(a+y)}{sina}$

132. Given, ${(x-a)}^2+{(y-b)}^2=c^2.$

Differentiating w r t 'x' we get

$\frac{d}{dx}{(x-a)}^2+\frac{d}{dx}{(y-b)}^2=\frac{d}{dx}{c}^2$

$\Rightarrow 2(x-a)+2(y-b)\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\frac{2(x-a)}{2(y-b)}=-\frac{(x-a)}{(y-b)}$

Again, $\frac{d^{2}y}{d{x}^2}=-\left\{\frac{(y-b)\frac{d}{dx}(x-a)-(x-a)\frac{d}{dx}(y-b)}{{(y-b)}^2}\right\}$

$=-\left\{\frac{(y-b)-(x-a)\frac{dy}{dx}}{{(y-b)}^2}\right\}$

$=-\left\{\frac{(y-b)+(x-a)\frac{(x-a)}{(y-b)}}{{(y-b)}^2}\right\}$

$=-\left\{\frac{{(y-b)}^2+{(x-a)}^2}{{(y-b)}^3}\right\}$

$=-\frac{c^2}{{(y-b)}^3}\left\{?{(x-a)}^2+{(y-b)}^2=c^2\right\}$

Then, L.H.S = $\frac{{\left\{1+{\left(\frac{dy}{dx}\right)}^2\right\}}^{3/2}}{\frac{d^{2}y}{d{x}^2}}=\frac{{\left\{1+\frac{{(x-a)}^2}{{(y-b)}^2}\right\}}^{3/2}}{\frac{-c^2}{{(y-b)}^2}}$

$=\frac{{\left\{{(y-b)}^2+{(x-a)}^2\right\}}^{3/2}}{{(y-b)}^3}*\frac{{(y-b)}^3}{-c^2}$

$=\frac{c^{2*3/2}}{-c^2}=\frac{c^3}{-c^2}=-c$ Where c is a constant and is independent of a and b.

131. Kindly go through the solution

130. Kindly go through the solution

129. Given, $y=12(1-cost). x=10(t-sint).$

Differentiating w r t 't' we get,

$\frac{dy}{dt}=12\frac{d}{dt}(1-cost)=12(0-(-sint))=12sint.$

$\frac{dx}{dt}=10\frac{d}{dt}(t-sint)=10(1-cost).$

$\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

$=\frac{12sint}{10(1-cost)}=\frac{12sint}{10[1-cost]}$

$=\frac{12*2sint/tcost/2}{10*2si{n}^{2}t/2}$

$\left\{\begin{array}{cc}\hfill ?sin2\theta =2sin\theta cos\theta \hfill & \hfill cos2\theta =1-2si{n}^2\theta \hfill \end{array}\right\}$ $=\frac{6}{5}\frac{cost/2}{sint/2}=\frac{6}{5}cott/2$

128. Let $y=x^{x^2-3}+{(x-3)}^{x^2}.$

Putting $u=x^{x^2-3} v={(x-3)}^{x^2}$ we get,

$y=u+v$

$\to \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$ ________(1)

Now, $u=x^{x^2-3}$

Taking log,

$logu=\left(x^2-3\right)logx.$

$\Rightarrow \frac{1}{u}\frac{du}{dx}=\left(x^2-3\right)\frac{d}{dx}logx+logx\frac{d}{dx}\left(x^2-3\right)$

$\Rightarrow \frac{du}{dx}=u\left[\frac{x^2-3}{x}+logx\left(2x\right)\right]$

$\frac{du}{dx}=x^{x^2-3}\left[\frac{x^2-3}{x}+2xlogx\right]$

And $v=(x-3)x^2$

$logv=x^{2}log(x-3)$

$\to \frac{1}{v}\frac{dv}{dx}=x^2\frac{d}{dx}log\left(x-3\right)+log(x-3)\frac{d}{dx}{x}^2$

$=\frac{x^2*1*}{x-3}\frac{d}{dx}(x-3)+log(x-3)\cdot 2x$

$\frac{dv}{dx}=v\left[\frac{x^2}{x-3}+2xlog(x-3)\right]$

$={(x-3)}^{x^2}\left[\frac{x^2}{x-3}+2xlog(x-3)\right]$

Hence eqn (1) becomes

$\frac{dy}{dx}=x^{\left(x^2-3\right)}\left[\frac{x^2-3}{x}+2xlogx\right]+{(x-3)}^{x^2}\left[\frac{x^2}{x-3}+2xlog(x-3)\right]$

127. Let $y=x^x+x^a+a^x+a^a.$

So, $\frac{dy}{dx}=\frac{d{x}^x}{dx}+\frac{d{x}^a}{dx}+\frac{d{a}^x}{dx}+\frac{d{a}^a}{dx}.$

$\to \frac{dy}{dx}=\frac{dy}{dx}+a{x}^{a-1}+a^{x}loga+0.$ _________(1)

Where $u=x^x$

$logu=xlogx$ (Taking log)

$\to \frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}logx+logx\frac{dx}{dx}$ (Differentiation w r t 'x')

$\to \frac{1}{y}\frac{dy}{dx}=\frac{x}{x}+logx$

$\to \frac{dy}{dx}=u[1+logx]$

$=x^x[1+logx]$

Hence eqn (1) becomes,

$\frac{dy}{dx}=x^x[1+logx]+a{x}^{a-1}+a^{x}loga.$

126. Let $y={(sinx-cosx)}^{sinx-cosx}$

Taking log,

$logy=(sinx-cosx)log(sinx-cosx)$

Differentiating w r t 'x' we get,

$\frac{1}{y}\frac{dy}{dx}=(sinx-cosx)\frac{dlog}{dx}(sinx-cosx)+log(sinx-cosx)\frac{d}{dx}(sinx-cosx).$

$=(sinx-cosx)*\frac{1}{(sinx-cosx)}*(cosx+sinx)+log(sinx-cosx)\left(cosx+sinx\right)$

$=(cosx+sinx)[1+log(sinx-cosx)]$

$\frac{dy}{dx}=y(cosx+sinx)[1+log(sinx-cosx)]$

$\frac{dy}{dx}={(sinx-cosx)}^{sinx-cosx}*(cosx+sinx)[1+log(sinx-cosx)]$

125. Let $y=cos(acosx+bsinx)$

So,  $\frac{dy}{dx}=-sin(acosx+bsinx)\frac{d}{dx}(acosx+bsinx)$

$=-sin(acosx+bsinx)[-asinx+bcosx].$

$=sin(acosx+bsinx)(asinx-bcosx)$