Solution:

`(x^2 + y^2)^2 = xy` ...(i)

Differentiating with respect to x, we have,

` 2 (x^2 + y^2) ( 2x + 2y .(dy)/(dx) ) = y + (xdy)/(dx)`

` => 4x(x^2 + y^2) + 4y(x^2 + y^2) .(dy)/(dx) = y + (xdy)/(dx)`

` => (dy)/(dx) (4x^2y + 4y^3 - x) = y - 4x^3 - 4xy^2`

` => (dy)/(dx) = ( y - 4x^3 - 4xy^2)/(4x^2y + 4y^3 - x)`

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Solution:

Given, `y = x^(x^x.......oo)`..............(i)

`=> y = x^y`

On taking log on both sides, we get

`log y= y log x`

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

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

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

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

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Solution:

` y = (sin)^x + sin^(-1) sqrtx`

Let `u = (sinx)^x` and `v = sin^(-1) sqrtx`

Now `y = u + v`

`(dy)/(dx) = (du)/(dx) + (dv)/(dx)` ....(i)

Consider `u = (sin x )^x`

Taking logarithms on both the sides, we have,

`logu = xlog (sinx)`

Differentiating with respect to x, we have,

` 1/u . (du)/(dx) = log (sin x) + x/(sin x) . cos x`

` => (du)/(dx) = (sin x)^x (log (sin x) + x cot x)` ...(ii)

Consider ` v = sin^(-1) sqrtx`

` (dv)/(dx) = 1/sqrt(1 -x) xx 1/(2sqrtx)` ....(iii)

From (i), (ii) and (iii)

We get, `(dy)/(dx) = (sin x)^x (log (sin x) +x cot x) + 1/(2 sqrtx sqrt(1-x))`

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Solution:

The given function is `( cos x)^y = (cos y)^x`

Taking logarithm on both the sides, we obtain

`y log cos x = x log cos y`

Differentiating both sides, we obtain

`log cos x xx (dy)/(dx) + y xx d/(dx) ( log cos x) xx d/(dx) (x) + x xx d/(dx) (log cos y)`

`=> log cosx xx (dy)/(dx) + y xx 1/(cosx) xx d/(dx) (cos x) = log cosy xx 1 + x xx 1/(cosy) xx d/(dx) (cosy)`

`=> log cosx xx(dy)/(dx) + y/(cos x) (- sin x) = log cos y + x/(cos y) xx (- sin y) xx (dy)/(dx)`

` => log cosx xx(dy)/(dx) - y tanx = log cosy - x tan y xx (dy)/(dx)`

` => log cosx xx(dy)/(dx) + x tan y xx (dy)/(dx) = log cosy - y tan x`

` => ( log cosx + x tany) xx (dy)/(dx) = log cosy - y tan x`

` :. (dy)/(dx) = ( log cosy + y tanx)/(log cosx + x tan y)`

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Solution:

`:. f(x) = log_e (log_e x)`

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

`f'(x) = 1/(log_e x) * 1/x`

`=> f' (e) = 1/(log_e e) * 1/e = 1/e = e^(-1)`

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Solution:

We have, `x^a y^b = (x- y)^(a + b)`

Taking log on both sides, we get

`a log x + b log y = (a+ b) log (x- y)`

Differentiating on both sides, we get

`a/x + b/y quad (dy)/(dx) = (a+ b) 1/ (x - y) ( 1 - (dy)/(dx))`

`=> (dy)/(dx) ( b/y + (a+b)/(x-y) )= (a+b)/(x-y) - a/x`

` => (dy)/(dx) = y/x . (bx +ay)/(bx +ay) = y/x => (dy)/(dx) - y/x = 0`

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Solution:

` y = a ( cos t + log tan t/2 )`

` =. (dy)/(dt) = a [ d/( dt) (cos t) + d/(dt) ( log tan t/2) ]`

` = a [ - sin t + cot t/2 xx sec^2 t/2 xx 1/2 ]`

` = a [ - sin t + 1/( 2 sin t/2 cos t/2 )]`

` = 2 ( - sin t + 1/( sin t) ) = a ( ( - sin^2 t + 1)/(sin t) ) = a (cos^2 t)/(sin t)`

`x = a sin t`

` (dx)/(dt) = a d/(dt) (sin t) = a cos t`

` :. (dy)/(dx) = ( (dy)/(dt))/( (dx)/(dt)) = ( a ( cos^2t)/(sin t) )/(a cos t) = (cos t)/(sin t) = cot t`

` (d^2y)/(dx^2) = - cosec^ t (dt)/(dx) = - cosec^ t xx 1/(a cos t) = - 1/( a sin^2 t cos t)`

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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:

`∵ 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 `f(x) = ax^2 + bx + c`

Then, `f( 1) = a + b + c`

and `f( -1) = a - b + c`

Since, `f( 1) = f( -1)`

`=> a + b + c = a - b + c`

`=> 2b = 0 => b = 0`

`:. f(x) = ax^2 + c`

`f (x) = ax^2 + c`

`f'(x) = 2ax`

`f'(a) = 2a (a) = 2a^2`

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Solution:

` y = a sin x + b cos x`

On differentiating with respect to `x`, we get

`(dy)/(dx) = a cos x - b sin x`

Now, `((dy)/(dx))^2 = ( a cos x - b sin x)^2`

`= a^2 cos^2 x + b^2 sin^2 1x`

`- 2 ab sin x cos x`

and `y^2 = (a sin x + b cos x)^2`

`= a^2 sin^2 x + b^2 cos^2 x`

`+ 2 ab sin x cos x`

So , ` ((dy)/(dx) )^2 + y^2 = a^2 (sin^2 x + cos^2 x)`

` + b^2 (sin^2 x + cos^2 x)`

` => ((dy)/(dx) )^2 + y^2 = a^2 + b^2 =` constant

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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`

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Solution:

`3^x + 3^y = 3^(x + y)`

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

` 3^x log3 + 3^y log 3 (dy)/(dx)`

`= 3^ (x + y) log 3 ( 1 + (dy)/(dx) )`

`=> 3^x + 3^y (dy)/(dx) = 3^(x + y) + 3^(x +y) (dy)/(dx)`

`=> (dy)/(dx) ( - 3^( x + y) + 3^y) = 3^(x+y) - 3^x`

` => (dy)/(dx) = ( 3^x (3^y - 1))/( 3^y (1 - 3^x)) = ( 3^(x - y) (3^y - 1))/(1 - 3^x)`

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Solution:

`e^y + xy = e`, when `x = 0`, then `y = 1`

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

`=> e^y (dy)/(dx) + y + x (dy)/(dx) = 0` ... (i)

At `x = 0, e (dy)/(dx) + 1 + 0 = 0 => (dy)/(dx) = - 1/e`

Again, differentiating Eq. (i) w.r.t. x, we get

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

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

At ` x = 0 , (d^2 y)/(dx^2) ( e + 0) + e (- 1/e)^2`

` + 2 ( - 1/e) = 0 `

`=> e (d^2 y)/(dx^2) - 1/e = 0 => (d^2 y)/(dx^2) = e^(-2)`

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Solution:

Given, `u = e^x sin x` and

`v = e^x cos x`

`(du)/(dx) = e^x ( sin x + cos x )` ,

` (dv)/(dx) = e^x (cos x - sin x)`

`(du)/(dx) = u + v` and `(dv)/(dx) = v - u`

` v (du)/(dx) - u (dv)/(dx) = v (u + v) - u (v - u)`

` => v (du)/(dx) - u (dv)/(dx) = u^2 + v^2`

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Solution:

We have, ` (dx)/(dt) = 1 - 1/t^2 , (dy)/(dt) = 1 + 1/t^2`

`:. (dy)/(dx) = (t^2 + 1)/(t^2 -1) = ( 1 + 2/(t^2 - 1))`

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

` = 2 . (-1)/( t^2 - 1)^2 . 2t xx t^2/(t^2 - 1) = (- 4 t^3)/( t^2 - 1)^3`

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Solution:

We have, `y = sin x + e^x`

` (dy)/(dx) = cos x + e^x => (dx)/(dy) = ( cos x + e^x)^(-1)`

` => (d^2 x)/(dy^2) = - (cos x + e^x)^(-2)`

` (- sin x + e^x) (dx)/(dy)`

` => (d^2x)/(dy^2) = (sin x - e^x)/(cos x + e^x)^2 . ( cos x + e^x)^(-1)`

` = (sin x - e^x)/(cos x + e^x)^3`

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Solution:

We have, `(dx)/(dy) = p => (dx)/(dy) = 1/p`

` :. (d^2 x)/(dy^2) = - 1/p^2 . (dp)/(dy) = 1/p^2 . d/(dy) ((dy)/(dx) )`

` = - 1/p^2 . (d^2 y)/(dx^2) . (dx)/(dy) = - q/p^3`

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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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