Ncert Solutions Maths class 12th

Given $f{x}^n$ is $y=e^x+1$

Differentiating with $x$ we get,

$y^{|}=\frac{dy}{dx}=e^x$

Again,

$y^{\left|\right|}=\frac{d^{2}y}{d{x}^2}=e^x$

Substituting value of $y^{\left|\right|}$ and $y^{|}$ in the given D.E. we get

$L.H.S.=y^{\left|\right|}-y^{|}=e^x-e^x=0=R.H.S$

$\therefore$ The given $f{x}^n$ is a solution of the given D.E.

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'

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

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

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

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

The highest order derivative present in the given D.E. is $\frac{d^{2}y}{d{x}^2}$ and its order is 2.

$\therefore$ Option (A) is correct.

In the given D.E,

$sin\frac{dy}{dx}$ is a trigonometric function of derivative $\frac{dy}{dx}$ . So it is not a polynomial equation so its derivative is not defined.

Hence, Degree of the given D.E. is not defined.

$\therefore$ Option (D) is correct.

The highest order derivative present in the D.E. is $y^{\left|\right|}$ so its order is 2.

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

The highest order derivative present in the D.E. is $y^{\left|\right|}$ so its order is 2.

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

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.

The given order derivative present in the D.E. is $y^{|}$ so its order is 1.

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

The highest order 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 derivative, its degree is 1.

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) $\left\{\frac{1}{1+1}+\frac{2*1}{1+1^2}+\frac{4*1^3}{1+1^4}+\frac{8*1^7}{1+1^8}\right\}$

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

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