Continuity and Differentiability

94. Kindly go through the solution

93. Kindly go through the solution

92. Kindly go through the solution

91. Given, $x=a\left(cost+logtan\frac{t}{2}\right)y=asint$

Differentiating w r t we get,

$\frac{dx}{dt}=a\frac{d}{dt}\left[cost+log\left(tan\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{1}{tan\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{d}{dt}\left(tan\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{1}{tan\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.se{c}^2\frac{t}{2}\frac{d}{dt}\left(\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{cos\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{sin\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}*\frac{1}{co{s}^2\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}*\frac{1}{2}\right]$

$=a\left[-sint+\frac{1}{2sin\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$=a\left[-sint+\frac{1}{sin2*\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$=a\left[-sint+\frac{1}{sint}\right]=a\left[\frac{1-si{n}^{2}t}{sint}\right]$

$=a\frac{co{s}^{2}t}{sint}\left\{?1=co{s}^{2}x+si{n}^{2}x\right\}$

$\frac{\mathrm{bd}y}{dt}=\frac{d}{dt}\left(asint\right)=acost$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}=\frac{acost}{\raisebox{1ex}{$aco{s}^{2}t$}\!\left/ \!\raisebox{-1ex}{$sint$}\right.}=\frac{sint}{cost}=tant$

90. Given, x = 4t and y = $\frac{4}{t}$ Differentiating w r t. 't' we get,

$\frac{dx}{dt}=4\frac{dy}{dt}=\frac{4d\left(t^{-1}\right)}{dt}$

$=-4t^{-2}$

$\frac{-4}{t^2}$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}=\frac{\raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$t^2$}\right.}{4}=-\frac{1}{t^2}.$

89. Given, x = sin t and y = cos2t. differentiation w r t. 't' we get,

$\begin{array}{l}\frac{dx}{dt}=cost\frac{dy}{dt}=\left(-sin2t\right)\frac{d2t}{dt}\\ -t\\ -2\left(2sintcost\right)\\ -tt\end{array}$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}$

$=\frac{-4sintcost}{cost}$

= -4 sin t

88. Kindly go through the solution

87. Given, x = 2at2 and y = at4. Differentiation w r t we get,

$\frac{dx}{dt}=4at.$ and $\frac{dy}{dt}=4a{t}^3.$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}=\frac{4a{t}^3}{4at}=t^2.$

86. To prove $\frac{d}{dx}(uv·w)=\frac{du}{dx}v·w+u·\frac{dv}{dx}·w+u·v·\frac{dw}{dx}.$

By repeating application of produced rule

$=\frac{d}{dx}(uv:u)$

$=u\frac{d}{dx}(u·w)+v·w\frac{dy}{dx}$

$u\left\{v\frac{dw}{dx}+w\frac{dv}{dx}\right\}+\frac{dy}{dx}v:w.$

$=u·v·\frac{dw}{dx}+u·\frac{dv}{dx}w+\frac{dy}{dx}·v·w$

$=\frac{dy}{dx}u·w+u·\frac{dv}{dx}·w+u·v·\frac{dw}{dx}$ = R*H*S*

By togarith differentiating,

Let y = u v w

Taking log, log y = log u + log v + log w

Differentiating w r t 'x'

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}.$

$\Rightarrow \frac{dy}{dx}=y\left[\frac{1}{4}\frac{dy}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}\right]$

$\Rightarrow \frac{d}{dx}(uvw)=uvw·\left[\frac{1}{u}\frac{dy}{dx}+\frac{1}{v}\frac{du}{dx}+\frac{1}{w}\frac{dur}{dx}\right]$

$=\frac{dy}{dx}v·w+u\frac{dv}{dx}·w+uv·\frac{dw}{dx}$

85. Let y = (x2- 5x + 8) (x3 + 7x + 9) ____ (1)

(i) by product rule

$\frac{dy}{dx}=\left(x^2-5x+8\right)\frac{d}{dx}\left(x^3+7x+9\right)+\left(x^3+7x+9\right)\frac{d}{dx}\left(x^2-5x+8\right)$

$=\left(x^2-5x+8\right)\left(3x^2+7\right)+\left(x^3+7x+9\right)(2x-5).$

3x4 + 7x2- 15x3- 35x + 24x2 + 56 + 2x4- 5x3 + 14x2- 35x 18x-45

= 5x4- 20x3 + 45x2- 52x + 11

(ii) $y=\left(x^2-5x+8\right)\left(x^3+7x+9\right)$

$y=x^5+7x^2+9x-5x^4-35x^2-45x+8x^3+56x+72$

$y=x^5-5x^4+15x^3-26x^2+11x+72.$

$\therefore \frac{dy}{dx}=5x^4-20x^3+45x^2-52x+11.$

Taking log in eqn (1)

$logy=log\left(x^2-5x+8\right)+log\left(x^3+7x+9\right)$

Now, Differe(iii) ntiating w r t 'x' we get,

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{x^2-5x+8}\frac{d}{dx}\left(x^2-5x+8\right)+\frac{1}{x^3+7x+9}\frac{d}{dx}\left(x^3+7x+9\right).$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}$

$\Rightarrow \frac{dy}{dx}=y\left[\frac{(2x-5)\left(x^3+7x+9\right)+\left(3x^2+7\right)\left(x^2-5x+8\right)}{\left(x^2-5x+8\right)\left(x^3+7x+9\right).}\right]$

$=\frac{y}{y}$ 2x4 + 14x4 + 18x- 35x- 45 + 3x1- 15x3 + 24x2 + 7x2- 35x + 56]

$\left\{?e{q}^n(1)\right\}.$

$\frac{dy}{dx}$ = 5x4- 20x3 + 45x2- 52x + 11

We observed that all the methods give the same result.