Class 12th

84. Given, f(x) = (1 + x)(1 + x 4)(1 + x 8)

Taking log,

logf(x) = log (1 + x) + log (1 + x) + log (1 + x 4) + log (1 + x 8)

Now, Differentiating w r t 'x' we get,

$\frac{1}{f(x)}{f}^{\prime }(x)=\frac{1}{1+x}\frac{d}{dx}(1+x)+\frac{1}{1+x^2}\frac{d}{dx}\left(1+x^2\right)+\frac{1}{1+x^4}\frac{d}{dx}\left(1+x^4\right)+\frac{1}{1+x^8}\frac{d\left(1+x^8\right)}{dx}$

$\Rightarrow f^{\prime }(x)=f(x)\left\{\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}\right\}.$

$\Rightarrow f^{\prime }(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left\{\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}\right\}$

Putting x = 1

f'(x) = (1 +1)(1 + 14)(1 +18) $\left\{\frac{1}{1+1}+\frac{2*1}{1+1^2}+\frac{4*1^3}{1+1^4}+\frac{8*1^7}{1+1^8}\right\}$

$=2*2*2+2\left\{\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right\}$

$=16*\left\{\frac{15}{2}\right\}=8*15=120.$

The highest order derivative present in the D.E. is $y^{\left|\right||}$ so its order is 3.

As the given D.E. is a polynomial equation in its derivation, its degree is 2.

$\frac{d^{4}y}{d{x}^4}$ As the given D.E. is a polynomial equation in its derivative, its degree is 1.

$\frac{d^{4}y}{d{x}^4}$ As the given D.E. is not a polynomial equation in its derivative, its degree is not defined.

83. Given, xy = ex-y.

Taking log,

log (x + y) = log (ex-y).

=logx + log y = (x-y) log e.

= logx +log y = x -y {Q log e = 1}

Differentiating w r t 'x' we get,

$\Rightarrow \frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}+\frac{dy}{dx}=1-\frac{1}{2}$

$\Rightarrow \frac{dy}{dx}\left(\frac{1}{y}+1\right)=\left(\frac{x-1}{x}\right)$

$\Rightarrow \frac{dy}{dx}\left(\frac{1+y}{y}\right)=\left(\frac{x-1}{x}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{y(x-1)}{x(1+y)}$

The highest order derivation present in the D.E. is $\frac{d^{2}s}{d{t}^2}$ so its order is 2 .

As the given D.E. is a polynomial equation in its derivative its degree is 1.

82. Given, (cos x)y = (cos y)x

Taking log, y log (cos x) = x log (cos y)

Differentiating w r t 'x' we get,

= $y\frac{d}{dx}$ log (cos x) + log (cos x) $\frac{dy}{dx}=x\frac{d}{dx}$ log (cos y) + dog (cos y) $\frac{dx}{dx}$

= y´ $\frac{1}{cosx}\frac{d}{dx}$ cos x + log (cos x) $\frac{dy}{dx}$ = x´ $\frac{1}{cosy}\frac{d}{dx}cosy+log(cosy).$

$\Rightarrow -y\frac{sinx}{cosx}+log(cosx)\frac{dy}{dx}=-x\frac{siny}{cosy}\frac{dy}{dx}+log(cosy)$

= log (cos x) $\frac{dy}{dx}$ + x tan $\frac{dy}{dx}=yctanx+log(cosy)$

$\Rightarrow \frac{dy}{dx}[log(cosx)+xtany]$ = y tan x + log (cos y )

$\Rightarrow \frac{dy}{dx}=\frac{ytanx+log(cosy}{log(cosx)+xtany}.$

The highest order derivation present in the differential equation (D.E.) is $\frac{d^{4}y}{d{x}^4}$ , so its order is 4.

As, the given D.E.is not a polynomial equation in its derivative, its degree is not defined.

81. Given, yx = xy

Taking log,

x log y .log x

Differentiating w r t 'x' we get,

$\Rightarrow x\frac{d}{dx}logy+logy\frac{dx}{dx}=y\frac{d}{dx}logx+logx\frac{dy}{dx}$

$\Rightarrow \frac{x}{y}·\frac{dy}{dx}+logy=\frac{y}{x}+logx\frac{dy}{dx}$

$\Rightarrow logx\frac{dy}{dx}-\frac{x}{y}\frac{dy}{dx}=logy-\frac{y}{x}$

$\Rightarrow \frac{dy}{dx}\left[\frac{ylogx-x}{y}\right]=\frac{xlogy-y}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{y(xlogy-y)}{x(ylogx-x)}.$