Solution:

`y = (1 + x^(1//4) ) (1 + x^(1//2) ) ( 1 - x^(1//4))`

`= (1 + x^(1//2) ) ( 1 - x^(1//2) ) = 1 - x`

`:. (dy)/(dx) = -1`

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Solution:

`y = sqrt ( cos x + sqrt (cos x + sqrt (cos x + sqrt (cos x + ..... oo ))))`

` :. y = sqrt (cos x + y) => y^2 = cos x + y`

On differentiating w.r.t. x, we get

` 2y (dy)/(dx) = - sin x + (dy)/(dx)`

` => sin x = (1 - 2 y) (dy)/(dx)`

` => (dy)/(dx) = (sin x)/(1 - 2y)`

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Solution:

According to question,

`|cos x|=-cos x`

`|sin x| =sin x`

`:. (dy)/(dx)=sin x+cos x`

`:. ((dy)/(dx))_((2n)/3) =sqrt 3/2-1/2=(sqrt 3-1)/2`

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Solution:

Let `y = sec (tan (sqrt x))`

By chain-rule,

`(dy)/(dx) = sec (tan sqrt(x) ) tan(tan sqrt(x) ) d/(dx) (tan sqrt(x) )`

`(dy)/(dx) = sec (tan sqrt(x) ) * tan (tan sqrt(x) ) sec^2 sqrt(x) * 1/(2* sqrt(x))`

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Solution:

`g(x)=f(x)/(1+f(x))`

`g^'(x)=((1+f(x))f^'(x)-f(x)f^'(x))/(1+f(x))^2`

`=(f^'(x))/(1+f(x))^2`

`f(x)=x/(x+1),f^'(x)=(1*(x+1)-x*1)/(x+1)^2=1/(x+1)^2`

`f(2)=2/3`

`f^'(2)=1/9`

`g^'(2)=(1/9)/(1+(2/3))^2=1/25`

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Solution:

When `x ge 0`, then `f(x) = |x |^3 = x^3`

`:. f'(x) = 3x^2` and `f''(x) = 6x`

which exists for all real values of `x`.

When `x <0 `, then `f'(x) = |x|^2 = (-x)^3 = -x^3`

`:. f'(x) = -3 x^2` and `f''(x) = -6x` which exist for all real values of `x`

Hence `f''(x) = {tt( (6x, text (if) x ge 0), (-6x, text(if) x < 0) )`

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Solution:

` :. y = gof (x) = g {f(x)}`

` = log (cos x)`

`:. (dy)/(dx) = 1/(cos x) (- sin x) = - tan x`

`=> ( (dy)/(dx))_(x = 0) = - tan 0 = 0`

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Solution:

`g(x) = f(e^x) e^(f(x))`

`:. g' (x) = f' (e^x) · e^x · e^(f (x))`

`+ f(e^x) · e^(f(x)) ·f ' (x)`

On putting `x = 0`,

`f (0) = f (1) = 0 , f' (1) = 2`, we get

`g' (0) = 2· 1 · 1 + 0 = 2`

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Solution:

By given condition, `phi (x) = f^(-1) (x)`

`:. f [ phi (x) ] = x`

On differentiating w.r.t. x, we get

`f ' [ phi (x) ] phi ' (x) = 1`

or `phi ' , (x ) = 1/(f' (phi (x))) = 1 + [ phi (x) ]^n `,

by definition of `f ' (x)`.

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Solution:

`f' (x) = x^2/x - kx + 1`

`=> f'(x) =x^2/x - ( (-5)/3) x + 1`

` = (x^ 2)/2 + 5/3 x + 1`

Now, differentiating both sides w.r.t. x, we get

` f"(x) = (2x)/3 + 5/3 = x + 5/3`

`:. f '' (-2/3) = -2/3 + 5/3 =1`

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Solution:

Let `x = cos 2 θ =>1/2 cos^(-1) x` ....(1)

` :. sqrt(1 +x) = sqrt( 1 + cos 2 theta) = sqrt( 1 + 2 cos^2 theta - 1) = sqrt2 cos theta`

` sqrt(1 -x) = sqrt( 1 - cos 2 theta) = sqrt( 1 - (1- 2 sin^2 theta )) = sqrt2 sin theta`

Let ` y = tan^(-1) [ (sqrt(1 + x) - sqrt (1 - x))/(sqrt(1 + x) + sqrt (1 - x)) ]`

` = tan^(-1) [ ( sqrt2 cos theta - sqrt2 sin theta)/(sqrt2 cos theta + sqrt2 sin theta)]`

` = tan^(-1) [ (1 - tan theta)/((1 + tan theta)]`

` = tan^(-1) { tan (pi/4 - theta ) }`

` = pi/4 - theta = pi/4 - 1/2 cos^(-1) x` ( From(1))

` :. (dy)/(dx) = - 1/2 ( (-1)/sqrt( 1 - x^2) ) = 1/(2 sqrt(1 - x^2))`

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Solution:

Given, `x = sqrt(a^(sin^(-1)t) )` .....(i)

and `y = sqrt(a^(cos^(-1)t) )` .....(ii)

On multiplying Eqs. (i) and (ii), we get

` xy = sqrt(a^(sin^(-1)t) ) xx sqrt(a^(cos^(-1)t) )`

` => xy sqrt( a^(sin^(-1)t) . a^(cos^-1) t)`

` = sqrt (a ^( sin^(-1) t + cos^(-1) t))`

` ( :. sin^(-1) x + cos^(-1) x = pi/2)`

` => xy = sqrt(a^(pi//12))`

On differentiating w.r.t. x, we get

` x (dy)/(dx) + y = 0 => x (dy)/(dx) = -y => (dy)/(dx) = (-y)/x \ \ \ \ ` [`:. d/(dx)`(constant) = 0]

`(dx)/(dy) = (-x)/y ``

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Solution:

Let `y = sin (tan^(-1) e^(-x) )`, diff. w.r.to `x`

`(dy)/(dx) = cos (tan^(-1) e^(-x) ) d/(dx) (tan^(-1) e^(-x))`

`= cos (tan^(-1) e^(-x) ) * 1/( 1+ e^(-2x)) * d/(dx) (e^(-x))`

`= - cos (tan^(-1) e^(-x)) 1/ (1+e^(-2x)) * e^(-x)`

---

Solution:

Let `y = sin^(-1) (( sqrt(1 + x) + sqrt( 1 - x))/2 )`

On putting `x = cos theta`, we get

` y = sin^(-1) ( 1/sqrt2 cos theta/2 + 1/sqrt2 sin theta/2 )`

`=> y = sin^(-1) { sin ( pi/4 + theta/2 ) }`

` => y = pi/4 + 1/2 theta`

` => y = pi/4 + 1/2 cos^(-1) x`

`:. (dy)/(dx) = - 1/(2 sqrt(1 - x^2))`

---

Solution:

`y = e^(sin^(-1) x), x = sin t`,

`:. y = e^t, (dt)/ (dx) =1/(sqrt (1-x^2)), (dy)/(dt) = e^t`

`:. (dy)/(dx) = (dy)/ (dt) * (dt)/ (dx) = e^t * 1/(sqrt (1-x^2)) = (e^(sin^(-1) x) )/(1- x^2)`

---

Solution:

`y= cos^(-1) ( (1-x^2)/ (1+x^2) )`, put `x= tan theta`,

`y = cos^(-1) ( (1-tan^(2) theta)/ (1+tan^(2) theta) ) = cos^(-1) (cos 2 theta) =2 theta`

`y = 2 tan^(-1) x :. (dy)/(dx) =2/(1+x^2)`

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Solution:

The given equation may be written as

`x sqrt (1+y)= -y sqrt (1+x)`


Squaring both sides, `x^(2) (1+y) = y^(2) (1+x)`

`=> x^2 - y^2=y^(2)x = x^(2)y`

`=>(x+y) (x-y) = -xy (x-y)`

`=> y= -x/ (1+x)`

`:. (dy)/(dx) = - { ( (1+x)*1 -x * (0+1) )/(1+x)^2} = -1/(1+x)^2`

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Solution:

The given equation is :

`2x^2-3xy+y^2+x+2y-8=0`

Differentiating this equation with respect to `x`, we get:

`4x-(3y+3xdy/dx)+2ydy/dx+1+2dy/dx-0=0`

`4x-3y-3xdy/dx+2ydy/dx+1+2dy/dx=0`

`rArr dy/dx=(3y-4x-1)/(2y-3x+2)`

---

Solution:

(i) `y = ae^x + be^(-x) + x^2`

`(dy)/(dx) = ae^(x) - be^(-x) + 2x`

`(d^2y)/(dx^2) = ae^x + be^(-x) + 2`

Now, `x (d^2y)/(dx^2) + 2 (dy)/(dx) -xy + x^2 - 2`

` =2x + 2 (ae^x - be^(-x)) + 4x - x^3 +x^2 - 2`

`=2 (ae^x - be^(-x)) - x^3 + x^2 + 6x - 2 != 0`

Hence, `y = ae^x + be^x + x^2` is not the solution of

the differential equation

`(d^2y)/(dx^2) + 2 - (dy)/(dx) - xy + x^2 - 2 = 0`

(ii) `y = e^x (a cos x + b sin x)`

` (dy)/(dx) = e^x [ - a sin x + b cos x] + e^x [ a cos x + b sin x ]`

` = e^x [(a + b) cos x - (a - b ) sinx ] `

` (d^2y)/(dx^2)= e^x [- 2a sin x + 2b cos x ]`

Now ` (d^2y)/(dx^2) - 2 (dy)/(dx) + 2y = e^x [(2a + 2a - 2b +2b) sin x + (2b - 2a - 2b + 2a) cosx] = 0`

Hence,` y = e^x (a cos x + b sin x)` is the solution

of the differential equation

` (d^2y)/(dx^2) + 2 (dy)/(dx) + 2y = 0`

(iii) `y = x sin 3x ,(dy)/(dx) = sin 3x + 3x cos 3x`

` (d^2y)/(dx^2) = 6 cos 3x - 9y`

` => (d^2y)/(dx^2) + 9y - 6 cos 3x = 0`

Hence, `y = x sin 3x` is the solution of

`(d^2y)/(dx^2) + 9y - 6 cos 3x = 0`

(iv) `x^2 = 2y^2 log y` ... (i)

Differentiating w.r.t. x

`2x = 2 [2y log y + y^2 xx 1/y ] (dy)/(dx)`

`= 2 [ 2y log y + y ] (dy)/(dx)`

`=> (dy)/(dx) = x/(2y log y + y) = x/( y (2 log y + 1))`

from (i) `2 log y = x^2/y^2`

`:. (dy)/(dx) = x/( y [x^2/y^2 + 1 ]) = (xy)/(x^2 + y^2)`

`=> (x^2 + y^2) (dy)/(dx) - xy = 0`

Hence, `x^2 = 2y^2 log y` is the solution of differential

equation `(x^2 + y^2) (dy)/(dx) - xy = 0`

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Solution:

We have, `x = sin t - tcos t`

and `y = t sin t + cos t`

`=> (dx)/(dt) = cos t - 1· cos t + t sin t = t sin t`

`=> (dy)/(dt) = 1· sin t + t cos t - sin t = t cos t`

` :. (dy)/(dt) = ( t cos t)/(t sin t) = cot t`

` ( (dy)/(dx))_(t = pi/2) = cot (pi/2) = 0`

---

Solution:

`\ frac{dx}{d\theta}=3\cos^2\theta(-\sin\theta)=-3\cos^2\theta\sin\theta`

and

`\ frac{dy}{d\theta}=3\sin^2\theta\cos\theta`

Hence

`1+ ( \ frac{dy}{dx} )^2=1+ ( \ frac{dy/(d\theta)}{dx/(d\theta)} )^2=1+\ frac{\sin^2\theta}{\cos^2\theta}=\sec^2\theta`

---

Solution:

Here `x= a (cos theta + theta sin theta )`

and `y = a (sin theta - theta cos theta)`

`(dx)/(d theta) = a [- sin theta + theta * cos theta + sin theta] = a theta cos theta`

`(dy)/(d theta) = a [cos theta + theta sin theta - cos theta] = a theta sin theta`

`(dy)/(dx) = ( (dy)/ (d theta)) /( (dx)/ (d theta)) = (a theta sin theta)/ (a theta cos theta) = tan theta`

---

Solution:

Let `y = log (log x)`

`(dy)/(dx) =1/(log x) * 1/x`

`(d^(2) y)/(dx^2) = 1/(log x) ( -1/ x^2) +1/x d /(dx) (1/(logx))`

`= -1/ (x^(2) log x) +1/x [ ( log x* 0 -1 * 1/x)/ (log x)^2 ]`

`= -1/(x^(2) log x) - 1/( x^2 (log x)^2)`

`= -1 /(x^(2) log x) [1+1/(log x)]`

---

Solution:

Let `y = sqrt ( ( (x-1)(x-2) )/ ( (x-3)(x-4)(x-5)) )`

Taking `log` on both sides, we get

`log y = log sqrt ( ( (x-1)(x-2) )/ ( (x-3)(x-4)(x-5)) )`

`log y =1/2 [ log (x-1 ) + log (x-2) -log (x -3)-log (x -4) -log (x- 5)]`

Differentiating w.r.t. `x` ,

`1/y (dy)/(dx) =1/2 [ 1/(x-1)+ 1/(x-2) -1/(x-3) -1/(x-4) -1/(x-5)]`

`:. (dy)/(dx) =1/2 sqrt ( ( (x-1)(x-2) )/ ( (x-3)(x-4)(x-5)) )


[ 1/(x-1)+ 1/(x-2) -1/(x-3) -1/(x-4) -1/(x-5)]`

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Solution:

Let `y = x^(1//x)`

At `x = e , y(e) = e^(1//e)`

`log y = 1/x log x`

`1/y . y_1 = 1/x . 1/x - 1/x^2 . log x`

`=> 1/y . y_1 = (1 - log x)/x^2 `

`=> y_1 = y ( (1 - log x)/x^2 )`

`=> y_2 = y_1 ( (1 - log x)/x^2 ) + y`

` [ ( x^2 (- 1/x) - ( 1 - log x) (2x) )/x^4 ]`

` = y_1 ( (1 - log x)/x^2 ) + ( y (- x - 2x + 2x log x))/x^4`

` = y_1 ( (1 - log x)/x^2 ) + ( y( 2x log x - 3x))/x^4`

` => y_2 (e) = 0 + ( e^(1//e) (2 e - 3 e))/e^4 = (- e^(1//e))/e^3`

` = -e^((1//e) - 3)`

---

Solution:

Let `f(x) = x^x`

`f'(x)=x^x [1+log x]`

`=x^x (log e +log x)`

`=x^x(log ex)`

Put `f'(x)=0 => x'(logex)=0`

`=> x=0,1//e`

Hence, function decreases in the interval `(0, 1// e)`.

---

Solution:

On taking log both sides, we get

`p log x + q log y = (p + q) log (x + y)`

`=> p 1/x + q 1/y (dy)/(dx)`

` = ( p + q) 1/(x +y) (1 + (dy)/(dx) )`

` p/x - (p + q)/(x + y) = ( (p + q)/(x + y) - q/y ) (dy)/(dx) `

` => ( py - qx)/(x (x + y)) = ( py - qx)/(y (x + y)) . (dy)/(dx) `

` :. (dy)/(dx) = y/x`

---

Solution:

Let `y = x^x+ x^a +a^x +a^a`,


Differentiating w.r.t. `x`,

`(dy)/(dx) = d/(dx) x^x +ax^(a-1) +a^(x) log a +0`

Put `u =x^x`


Taking `log` on both sides

`log u = x log x, 1/u (du)/(dx) = 1* log x + x* 1/x`

`=> (du)/(dx) = u (1+ log x) = x^x (1+ log x)`

`:. (dy)/(dx) = x^x (1+ log x) +ax^(a-1) + a^(x) log a`

---

Solution:

Let `u =` In `(x + sin x)` and `v = x + cos x`

Now, `(du)/(dx) = 1/(x +sin x) ( 1 + cos x)`

and ` (dv)/(dx) = 1- sin x`

Now, we can find derivative ofu w.r.t. v, we get

`(du// dx)/ (dv//dx) = ( (1 + cos x)// (x + sin x) )/(1-sin x)`

` => (du)/(dv) = (1 + cos x)/( (x + sin x) (1 - sin x))`

---

Solution:

Let `t = logcosx`

`f(t) = e^(sint)`

` f'(t) = (df(t)) / (dt) = e^(sint) cost`

`g(t) = t`

`g'(t) = 1`

On dividing both `(f'(t)) / (g'(t)) = e^(sint) cost = f(x).cos(g(t)) `

Now replace `t` with `x`

`(f'(x)) / (g'(x)) = e^(sinx) cosx = f(x).cos(g(x)) `


Alternatively :

`:. f(x) = e^(sin (log cos x))`

`:. f'(x) = e^(sin (log cos x) ) * cos (log cos x) * 1/(cos x) (- sin x)`

`= -e^(sin (log cos x) ) * cos (log cos x) * tan x`

and `g(x) = log cos x`

`:. g'(x) = 1/(cos x) (- sin x)= - tan x`

Hence, `(f'(x))/(g'(x)) = (-e^(sin (log cos x) ) * cos (log cos x) * tan x)/(- tan x)`

`= e^(sin (log cos x) ) * cos (log cos x)`

`= f(x) * cos [g(x)]`

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Solution:

Let `u = log_x 5` and `v = log_5 x`

Then, `(du)/(dx) = (-log 5)/(log x)^2 * 1/x ` and `(dv)/(dx) = 1/(x log 5)`

`(du)/(dv) = ((du)/(dx))/((dv)/(dx)) = ( (-log 5)/(log x)^2 * 1/x ) /((1/(log 5)) * 1/x) = - ((log 5)/(log x))^2 = - (log_x 5)^2 `

---

Solution:

We have, `y = e^(2x)`

`=> (dy)/(dx) = e^(2x) . 2`

` => (d^2 y)/(dx^2) = 4e^(2x)`

Now, `log y = 2x, x = 1/2 log y`

` (dy)/(dx) = 1/(2y) , (d^2 x)/(dy^2) = - 1/(2y^2) = - 1/(2e^(4x))`

`:. (d^2 y)/(dx^2) . (d^2 x)/(dy^2) = 4e^(2x) xx ( - 1/(2e^(4x)) )`

` = - 2e^(-2x)`

---

Solution:

`∵ y^2 = P(x)` ... (i)

`2y (dy)/(dx) = P ' (x)` ......(ii)

and ` 2y (d^2y)/(dx^2) + 2 ( (dy)/(dx))^2 = P '' (x)`

`=> 2y (d^2y)/(dx^2) = y^2 P '' (x) - 2 ( (dy)/(dx))^2`

` => 2y^3 (d^2y)/(dx^2) = y^2 P '' (x) - 2 ( y (dy)/(dx))^2`

` => 2y^3 (d^2y)/(dx^2) = P (x) P '' (x) - 2 { (P'' (x) )/2 }^2`

[using Eqs. (i) and (ii)]

` => 2y^3 (d^2y)/(dx^2) = P (x) P '' (x) - 1/2 { P' (x) }^2`

`=> 2 d/(dx) { y^3 (d^2y)/(dx^2) }`

` = P (x) P ''' (x) + P ' (x) P '' (x)`

`- 1/2 . 2 P ' (x) '' (x)`

` = P (x) P ''' (x)`

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Solution:

`tan^(-1) ((1 - 2 log x)/( 1 + 2 log x))`

` + tan^(-1) ((3 + 2 log x)/( 1 - 3. 2 log x))`

`= tan^(-1) 1 - tan^(-1) ( 2 log x)`

`+ tan^(-1) 3 + tan^( - 1) (2 log x)`

`= tan^(-1) 1 + tan^(-1 ) 3`

`:. y =` constant

` :. (dy)/(dx) = 0` and ` (d^2 y)/(dx^2) = 0`

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Solution:

Let `y = x^(1//x)`

At `x = e , y(e) = e^(1//e)`

`log y = 1/x log x`

`1/y . y_1 = 1/x . 1/x - 1/x^2 . log x`

`=> 1/y . y_1 = (1 - log x)/x^2 `

`=> y_1 = y ( (1 - log x)/x^2 )`

`=> y_2 = y_1 ( (1 - log x)/x^2 ) + y`

` [ ( x^2 (- 1/x) - ( 1 - log x) (2x) )/x^4 ]`

` = y_1 ( (1 - log x)/x^2 ) + ( y (- x - 2x + 2x log x))/x^4`

` = y_1 ( (1 - log x)/x^2 ) + ( y( 2x log x - 3x))/x^4`

` => y_2 (e) = 0 + ( e^(1//e) (2 e - 3 e))/e^4 = (- e^(1//e))/e^3`

` = -e^((1//e) - 3)`

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