Class 12th

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

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

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

124. Let $y={(logx)}^{logx}$

Taking log,

$logy=logxlog(logx)$

Differentiating w r t. x, we get,

$\frac{1}{y}\frac{dy}{dx}=logx\frac{d}{dx}log(logx)+log(logx)\frac{d}{dx}(logx)$

$=logx*\frac{1}{logx}\frac{d}{dx}logx+\frac{log(logx)}{x}$

$=\frac{1}{x}+\frac{log(logx)}{x}$

$\frac{dy}{dx}=\frac{y}{x}[1+log(logx)].$

$={(logx)}^{logx}\left[\frac{1+log(logx)}{x}\right]$

123. Kindly go through the solution

122. Kindly go through the solution

121. Kindly go through the solution

In (D), each of the terms has a degree 2.

Hence, (D) is homogenous

$\therefore$ Option (D) is correct.