Continuity and Differentiability

111. Given, $y={\left(ta{n}^{-1}x\right)}^2$

So, $y_1=\frac{dy}{dx}=2\left(ta{n}^{-1}x\right)\frac{d}{dx}ta{n}^{-1}x$

$\Rightarrow y_1=2ta{n}^{-1}x*\frac{1}{1+x^2}$

$\left(x^2+1\right)y_1=2ta{n}^{-1}x$

Differentiating again w r t 'x' we get,

$\left(x^2+1\right)\frac{d{y}_1}{dx}+y_1\frac{d}{dx}\left(x^2+1\right)=2\frac{d}{dx}ta{n}^{-1}x$

$\Rightarrow \left(x^2+1\right)y_2+y_1\left(2x\right)=\frac{2}{1+x^2}$

$\Rightarrow {\left(x^2+1\right)}^2{y}_2+2x\left(1+x^2\right)y_1=2$

Hence proved.

110. Kindly go through the solution

109. Given,  $y=500{e^7}^x+600{e^{-7}}^x$

So,  $\frac{dy}{dx}=500*7{e^7}^x+600\left(-7\right){e^{-7}}^x$

$\therefore \frac{d^{2}y}{d{x}^2}=500*7^2{e^7}^x+600*7^2{e^{-7}}^x$

$=49\left[500e^{7x}+600e^{-7x}\right]$

$=49*y$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=49y$

108. Given, $y=A{e}^{mx}+B{e}^{nx}$ _______(1)

So, $\frac{dy}{dx}=Am{e}^{mx}+Bn{e}^{nx}$ _______(2)

$\therefore \frac{d^{2}y}{d{x}^2}=A{m}^2{e}^{mx}+B{n}^2{e}^{nx}$ _________(3)

So, L.H.S = $\frac{d^{2}y}{d{x}^2}-\left(m+n\right)\frac{dy}{dx}+mny$

$=A{m}^2{e}^{mx}+B{n}^2{e}^{nx}-\left(m+n\right)\left[Am{e}^{mx}+Bn{e}^{nx}\right]+mn\left[A{e}^{mx}+B{e}^{nx}\right]$

$=A{m}^2{e}^{mx}+B{n}^2{e}^{nx}-A{m}^2{e}^{mx}-Bmn{e}^{nx}-Amn{e}^{mx}-B{n}^2{e}^{nx}+Amn{e}^{mx}+Bmn{e}^{nx}$

= 0 = R.H.S.

107. Given $y=3cos\left(logx\right)+4sin\left(logx\right)$

So, $y_1=\frac{dy}{dx}=3\frac{d}{dx}cos\left(logx\right)+4\frac{d}{dx}sin\left(logx\right)$

$y_1=3\left[-sin\left(logx\right)\right]\frac{d}{dx}logx+4cos\left(logx\right)\frac{d}{dx}\left(logx\right)$

$y_1=\frac{-3sin\left(logx\right)}{x}+\frac{4cos\left(logx\right)}{x}$

$\Rightarrow x{y}_1=-3sin\left(logx\right)+4cos\left(logx\right)$ ______________(1)

Differentiating eqn (1) w r t 'x' we get,

$\frac{d}{dx}\left(x{y}_1\right)=-3\frac{d}{dx}sin\left(logx\right)+4\frac{d}{dx}cos\left(logx\right)$

$\Rightarrow x\frac{d{y}_1}{dx}+y_1\frac{dx}{dx}=-3cos\left(log\left(x\right)\right)\frac{d}{dx}logx+4\left[-sin\left(logx\right)\right]\frac{d}{dx}logx$

$\Rightarrow x{y}_2+y_1=\frac{-3cos\left(logx\right)}{x}-\frac{4sin\left(logx\right)}{x}$

$\Rightarrow x^2{y}_2+y_1=-\left[3cos\left(logx\right)+4sin\left(logx\right)\right]$

$\Rightarrow x^2{y}_2+y_1=-y$

$\Rightarrow x^2{y}_2+y_1+y=0$

106. Kindly go through the solution

105. Given,  $y=5cosx-3sinx$

Differentiating w r t x we get,

$\frac{dy}{dx}=5\frac{d}{dx}cosx-3\frac{d}{dx}sinx$

$-5sinx-3cosx.$

Differentiating again w r t. 'x' we get,

$\frac{d^{2}y}{d{x}^2}=-5\frac{d}{dx}sinx-3\frac{d}{dx}cosx$

$=-5cosx+3sinx$

$=-\left[5cosx-3sinx\right]$

$=-y$

$\Rightarrow \frac{d^{2}y}{d{x}^2}+y=0$ . Hence proved.

104. Let $y=sin\left(logx\right)$

so, $\frac{dy}{dx}=\frac{d}{dx}sin\left(logx\right)=cos\left(logx\right)\frac{d}{dx}logx=\frac{cos\left(logx\right)}{x}$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{x\frac{d}{dx}cos\left(logx\right)-cos\left(logx\right)\frac{dx}{dx}}{x^2}$

$=\frac{x\left[-sin\left(logx\right)\right]\frac{d}{dx}logx-cos\left(logx\right)}{x^2}$

$=\frac{-\left[x\cdot sin\left(logx\right)*\frac{1}{x}+cos\left(logx\right)\right]}{x^2}$

$=\frac{-\left[sin\left(logx\right)+cos\left(logx\right)\right]}{x^2}$

103. Let $y=log\left(logx\right)$

So,  $\frac{dy}{dx}=\frac{1}{logx}\frac{d}{dx}logx=\frac{1}{xlogx}$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{xlogx\frac{d}{dx}\left(1\right)-1\cdot \frac{d}{dx}\left(xlogx\right)}{{\left(xlogx\right)}^2}$

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

$=\frac{-\left(x*\frac{1}{x}+logx\right)}{{\left[xlogx\right]}^2}$

$=\frac{-\left(1+logx\right)}{{\left(xlogx\right)}^2}$

102. Let $y=ta{n}^{-1}x$

So,  $\frac{dy}{dx}=\frac{d}{dx}ta{n}^{-1}x=\frac{1}{1+x^2}$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{\left(1+x^2\right)\frac{d}{dx}\left(1\right)-\left(1\right)\frac{d}{dx}\left(1+x^2\right)}{{\left(1+x^2\right)}^2}$

$=\frac{-2x}{{\left(1+x^2\right)}^2}$