Continuity and Differentiability

$f\text{'}\left(x\right)=\frac{n_1\cdot f\left(x\right)}{x-3}+n_2\cdot \frac{f\left(x\right)}{x-5}$

$=\frac{f\left(x\right)\cdot \left(n_1+n_2\right)}{\left(x-3\right)\left(x-5\right)}\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_2}\right)$

$f\text{'}\left(x\right)={\left(x-3\right)}^{n_1-1}\cdot {\left(x-5\right)}^{n_2-1}\cdot \left(n_1+n_2\right)\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_{2l}}\right)$  

option (C) is incorrect, there will be minima.

$Inx\in \left(-2,2\right)$  

 |x|- 1| is not differentiable at x = -1, 0, 1

|cospx| is not differentiable at x =  $\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2}$ -  

(a + √2bcosx) (a - √2bcosy) = a² - b²
⇒ a² - √2abcosy + √2abcosx - 2b²cosxcosy = a² - b²
Differentiating both sides:
0 - √2ab (-siny dy/dx) + √2ab (-sinx) - 2b² [cosx (-siny dy/dx) + cosy (-sinx)] = 0
At (π/4, π/4):
ab dy/dx - ab - 2b² (-1/2 dy/dx + 1/2) = 0
⇒ dy/dx = (ab+b²)/ (ab-b²) = (a+b)/ (a-b); a, b > 0

For x>2, f (x) = ∫? ¹ (5+1-t)dt + ∫? ² (5+t-1)dt + ∫? (5+t-1)dt
= ∫? ¹ (6-t)dt + ∫? ² (4+t)dt + ∫? (4+t)dt
= [6t-t²/2]? ¹ + [4t+t²/2]? ² + [4t+t²/2]?
= (6-1/2) + (8+2 - (4+1/2) + (4x+x²/2 - (8+2)
= 5.5 + 5.5 + 4x+x²/2 - 10 = 4x+x²/2 + 1.
f (2? ) = 8+2+1 = 11. f (2? ) = 5 (2)+1 = 11. Continuous.
f' (x) = 4+x for x>2. f' (2? ) = 6.
For x<2, f' (x)=5. f' (2? )=5.
Not differentiable at x=2.

81. Given, yx = xy

Taking log,

x log y .log x

Differentiating w r t 'x' we get,

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

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

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

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

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

80. Given, xy + yx = 1

Let 4 = xy and v =., we have,

u + v = 1.

$\Rightarrow \frac{dy}{dx}+\frac{dv}{dy}=0$ ___ (1)

So, u = xy

= log u = y log x(taking log)

Now, differentiating w r t 'x',

$\frac{1}{4}\frac{dy}{dx}=y\frac{d}{dx}logx+logx\frac{dy}{dx}.$

$\frac{dy}{dx}=4\left[\frac{y}{x}+logx\frac{dy}{dx}\right]$

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

= xy- 1y + xy log x $\frac{dy}{dx}.$

And v = yx.

log v = x log y.

Differentiating w r t 'x',

$\Rightarrow \frac{1}{v}\frac{dv}{dx}=x\frac{d}{dx}logy+logy\frac{dx}{dx}$

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

$\Rightarrow \frac{dv}{dx}=v\left[\frac{x}{y}\frac{dy}{dx}+logy\right]$

$=y^x·\frac{x}{y}·\frac{dy}{dx}+y^{x}log·y$

= yx- 1. $x\frac{dy}{dx}$ + yx log y.

So, eqn (1) becomes

xy- 1y + xy log x $\frac{dy}{dx}$ + yx - 1 $\frac{dy}{dx}$ + yx log y = 0

$\Rightarrow \frac{dy}{dx}\left(x^{y}logx+y^{x+1}·x\right)$ = - (xy- 1y + yx log y)

$\Rightarrow \frac{dy}{dx}=\frac{-\left(x^{y-1}·y+y^{x}logy\right)}{\left(x^{y}logx+y^{x-1}·x\right).}$

78. Let y = x^x cos x  $\frac{a^2+1}{x^2-1}$

Putting  4 = x^x cos x and v = $\frac{x^2+1}{x^2-1}$ we have,

y = u + v

$\Rightarrow \frac{dy}{dx}=\frac{dy}{dx}+\frac{dv}{dx}$ ____ (1)

As u $\left|=\right|$ x^x cos x.

Taking log,

Log u = x cos x log x

Differentiating w r t 'x',

$\frac{1}{u}\frac{dy}{dx}=x\frac{d}{dx}$ [cos x log x] + cos x log x $\frac{dx}{dx}$

= x $\left\{cotx\frac{d}{dx}logx+logx\frac{d}{dx}cosx\right\}$ + cos x log x.

$=x\left\{cosx·\frac{1}{x}-sinx·logx\right\}$ + cos x log x.

= cos x- sin x. log x + cos x log x.

$\frac{dy}{dx}=u$ [cosx + cos x log x- sin x log x]

= x^x cos x [cos x + cos x log x-x sin x log x]

And v = $\frac{x^2+1}{x^2-1}$

So, $\frac{dv}{dx}=\frac{\left(x^2-1\right)\frac{d}{dx}\left(x^2+1\right)-\left(x^2+1\right)\frac{d}{dx}(x-1)}{{\left(x^2-1\right)}^2}$

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

$=\frac{2x^3-2x-2x^3-2x}{{\left(x^2-1\right)}^2}$

$=\frac{-4x}{{\left(x^2-1\right)}^2}.$

Hence, eqn (1) becomes,

$\frac{dy}{dx}$ x^{xcos x} [cos x + cos x log x-x sin x log x] $\frac{-4x}{{\left(x^2-1\right)}^2}$

76. Kindly go through the solution