Continuity and Differentiability

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

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

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

101. Let $y=e^{6x}cos3x$

So, $\frac{dy}{dx}=e^{6x}\frac{d}{dx}cos3x+cos3x\frac{d}{dx}{e}^{6x}$

$=e^{6x}\left(-sin3x\right)\frac{d}{dx}\left(3x\right)+cos3x\cdot e^{6x}\frac{d}{dx}\left(6x\right)$

$=e^{6x}\left[-3sin3x+6cos3x\right]$

$\therefore \frac{d^{2}y}{d{x}^2}=e^{6x}\frac{d}{dx}\left[-3sin3x+6cos3x\right]+\left[-3sin3x+6cos3x\right]\frac{d}{dx}{e}^{6x}$

$=e^{6x}\left[-3cos3x\frac{d}{dx}\left(3x\right)+6\left(-sin3x\right)\frac{d}{dx}\left(3x\right)\right]+\left[-3sin3x+6cos3x\right]e^{6x}\frac{d}{dx}\left(6x\right)$

$=e^{6x}\left\{-9cos3x-18sin3x-18sin3x+36cos3x\right\}$

$=e^{6x}\left(27cos3x-36sin3x\right)$

$=9e^{6x}\left(3cos3x-4sin3x\right)$

100. Let $y=e^{x}sin5x$

so, $\frac{dy}{dx}=e^x\frac{d}{dx}sin5x+sin5x\frac{d}{dx}{e}^x$

$=e^x\cdot cos5x\frac{d}{dx}\left(5x\right)+e^{x}sin5x$

$=5e^{x}cos5x+e^{x}sin5x.$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(5e^{x}cos5x+e^{x}sin5x\right)$

$=5e^x\frac{d}{dx}cos5x+5cos5x\frac{d}{dx}{e}^x+e^x\frac{d}{dx}sin5x+sin5x\frac{d}{dx}{e}^x$

$=-5e^{x}sin5x\frac{d}{dx}\left(5x\right)+5e^{x}cos5x+e^{x}cos5x\frac{d}{dx}\left(5x\right)+e^{x}sin5x$

$=-25e^{x}sin5x+5e^{x}cos5x+5e^{x}cos5x+e^{x}sin5x$

$=e^x\left(10\cdot cos5x-24sin5x\right)$

$=2e^x\left(5cos5x-12sin5x\right)$

99. Let $y=x^{3}logx$

So,  $\frac{dy}{dx}=x^3\frac{d}{dx}logx+log2.\frac{d{x}^3}{dx}$

$=x^3.\frac{1}{x}+logx\cdot 3x^2$

$=x^2+logx\left(3x^2\right)$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(x^2+logx\cdot 3x^2\right)$

$=2x+6xlogx+3x^2*\frac{1}{x}$

$=2x+6xlogx+3x$

$=5x+6xlogx$

98. Let $y=logx$

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

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\frac{1}{x}=\frac{d}{dx}{x}^{-1}=-1x^{-1-1}=\frac{-1}{x^2}$

97. Let $y=xcosx$

So,  $\frac{dy}{dx}=x\frac{d}{dx}cosx+\frac{dx}{dx}cosx$

$=-xsinx+cosx$

$\therefore \frac{d^{2}y}{d{x}^2}=-x\frac{d}{dx}sinx-sinx\frac{dx}{dx}+\frac{d}{dx}cosx$

$=-xcosx-sinx+\left(-sinx\right)$

$=-\left(xcosx+2sinx\right)$

96. Let $y=x^{20}$

So,  $\frac{dy}{dx}=20x^{20-1}=20x^{19}$

$\therefore \frac{d^{2}y}{d{x}^2}=20*19x^{19-1}$

$=380x^{18}$

95. Let y = x2 + 3x + 2

So,  $\frac{dy}{dx}=2x+3+0$  (differentiation w r t 'x')

$?\frac{d^{2}y}{d{x}^2}=2+0$  (Again “ “ ) = 2