Q: If x 1+y +y 1+x =0, then d y d x =

aligned & Given, x 1+y +y 1+x =0 \\ & x 1+y =-y 1+x \\ & (x 1+y )^2=(-y 1+x )^2 aligned array lr & x^2(1+y)=y^2(1+x) \\ & x^2-y^2=y^2 x-x^2 y \\ & (x-y)(x+y)=x y(y-x) \\ & x+y=-x y \\ & x+y+x y=0 \\ & y= -x 1+x =-1+ 1 1+x array Differentiating w.r.t, x d y d x =- 1 (1+x)^2

Q: If y= ^ -1 ( x -x 1+x^ 3 2 ), then y^ (1) is equal to

Given, aligned y & = ^ -1 ( x -x 1+x^ 3 2 ) \\ y & = ^ -1 ( x -x 1+ x x ) \\ y & = ^ -1 ( x )- ^ -1 x \\ & [ ^ -1 ( x-y 1+x y )= ^ -1 x- ^ -1 y ] aligned Differentiating w.r.t x y^ (x)= d y d x = 1 1+x 1 2 x - 1 1+x^2 Now, y^ (1)= 1 2 1 2 - 1 2 =- 1 4

Q: If 2^ x +2^ y =2^ x+y , then d y d x =

On differentiating array l 2^ x 2+2^ y 2 dy dx \\ =2^ x 2^ y dy dx 2+2^ y 2^ x 2 \\ 2^ x +2^ y dy dx =2^ x+y dy dx +2^ x+y \\ dy dx = 2^ x+y -2^ x 2^ y -2^ x+y \\ dy dx = 2^ x +2^ y -2^ x 2^ y -2^ x -2^ y =-2^ y-x array

Question 1

If x 1+y +y 1+x =0, then d y d x =

Accepted Answer

aligned & Given, x 1+y +y 1+x =0 \ & x 1+y =-y 1+x \ & (x 1+y )^2=(-y 1+x )^2 aligned array lr & x^2(1+y)=y^2(1+x) \ & x^2-y^2=y^2 x-x^2 y \ & (x-y)(x+y)=x y(y-x) \ & x+y=-x y \ & x+y+x y=0 \ & y= -x 1+x =-1+ 1 1+x array Differentiating w.r.t, x d y d x =- 1 (1+x)^2

Question 2

If y= ^ -1 ( x -x 1+x^ 3 2 ), then y^ (1) is equal to

Accepted Answer

Given, aligned y & = ^ -1 ( x -x 1+x^ 3 2 ) \ y & = ^ -1 ( x -x 1+ x x ) \ y & = ^ -1 ( x )- ^ -1 x \ & [ ^ -1 ( x-y 1+x y )= ^ -1 x- ^ -1 y ] aligned Differentiating w.r.t x y^ (x)= d y d x = 1 1+x 1 2 x - 1 1+x^2 Now, y^ (1)= 1 2 1 2 - 1 2 =- 1 4

Question 3

At x= ^2 4 , d d x ( ^ -1 ( x )+ ^ -1 (e^x ) )=

Accepted Answer

d d x ( ^ -1 ( x )+ ^ -1 ( e ^x ) ) = (- x ) 1+ ^2 x ( 1 2 x )+ 1 e^x e^ 2 x -1 e^x but when x= ^2 4 =- - 2 1+ ^2 2 ( 1 2 ) ( 2 )+ 1 e^ ^2 / 2 -1 =- 1 + 1 e^ ^2 / 2 -1

Question 4

The slope of the tangent to the curve x=3 t^2+1, y=t^3-1 at x=1 is:

Accepted Answer

Given curve is x=3 t^2+1 Second curve is y=t^3-1 d y d x = d y d t d t d x =3 t^2 1 6 t = t 2 But from (i) when x=1 we have 1=3 t^2+1 3 t^2=0 t=0 When x =1 then t =0 dy dx = 0 Hence, slope of the tangent to the curve =0

Question 5

If y^x=e^ y-x , then d y d x is equal to

Accepted Answer

Here, y ^ x = e ^ y - x Taking log on both sides, we get aligned & y^x= e^ y-x \ & ( a^b=b a and e=1 ) aligned x y=(y-x) e x y=y-x (i) On differentiating w.r.t. x, we get d d x (x y)= d d x (y-x) (using product rule) aligned & x ( 1 y ) d y d x + y(1)= d y d x -1 \ & d y d x ( x y -1 )=-1- y \ & d y d x [ y (1+ y) y -1 ]=-(1+ y) \ & [ from eq. (i) x= y (1+ y) ] \ & d y d x [ 1-1- y 1+ y ]=-(1+ y) aligned d y d x =- (1+ y)^2 - y d y d x = (1+ y)^2 y

Question 6

If f ( x )= 1+ ^2 ( x ^2 ) , then the value of f^ ( 2 )

Accepted Answer

On differentiating (i) w.r.t.x, we get f^ (x)= -2 x^2 x^2 1+ ^2 x^2 (x) Put, x= 2 in (ii), we get aligned & f ^ ( 2 )=- 2 2 ( 4 ) 1+ 1 2 \ & =- 2 2 3 2 =- 6 aligned

Question 7

The derivative of e^ x^3 with respect to x is

Accepted Answer

Let y= e ^ x ^3 , z = x On differentiating w.r.t.x, we get d y d x =e^ x^3 (3 x^2 )=3 x^2 e^ x^3 and d z d x = 1 x d y d z = d y d x d z d x = 3 x^2 e^ x^3 ( 1 x ) =3 x^3 e^ x^3

Question 8

If y= ( 1+ 2 1- 2 ) , then d y d at = 3 4 is :

Accepted Answer

y= 1+ 2 1- 2 y= 2 ^2 2 ^2 = ^2 y= Differentiate w.r.t. ' ', we get: d y d =- cosec ^2 Now, ( dy d )_ = 3 4 =- cosec ^2 ( 3 4 ) =- cosec ^2 ( - 4 )=- cosec ^2 4 =-2

Question 9

If 2^ x +2^ y =2^ x+y , then d y d x =

Accepted Answer

On differentiating array l 2^ x 2+2^ y 2 dy dx \ =2^ x 2^ y dy dx 2+2^ y 2^ x 2 \ 2^ x +2^ y dy dx =2^ x+y dy dx +2^ x+y \ dy dx = 2^ x+y -2^ x 2^ y -2^ x+y \ dy dx = 2^ x +2^ y -2^ x 2^ y -2^ x -2^ y =-2^ y-x array

Question 10

If x= 1-t^ 2 1+t^ 2 and y= 2 t 1+t^ 2 , then d y d x is equal to :

Accepted Answer

Let x= 1-t^ 2 1+t^ 2 and y= 2 t 1+t^ 2 Put t= , we get x= 1- ^ 2 1+ ^ 2 and y= 2 1+ ^ 2 x= 2 and y= 2 dx d =-2 2 and dy d =2 2 Now, dy dx = dy d d dx =- 2 2 =- x y

Question 11

d d x ( ^ -1 1+ x 2 1- x 2 ) is equal to

Accepted Answer

We have d d x ( ^ -1 1+ x 2 1- x 2 ) = d d x [ ^ -1 1+ (2 ^ 2 x 4 -1 ) 1- (1-2 ^ 2 x 4 ) ] = d d x ( ^ -1 . 2 ^ 2 x 4 2 ^ 2 x 4 ) . = d d x ( ^ -1 ^ 2 x 4 )= d d x ( ^ -1 x 4 ) = d d x ( ^ -1 ( 2 - x 4 ) )= d d x ( 2 - x 4 )=0- 1 4 =- 1 4

Question 12

If x y+y^ 2 = x+y, then find d y d x is

Accepted Answer

The given relation is x y+y^ 2 = x+y. Differentiating both sides with respect to x, we get d d x (x y)+ d d x (y^ 2 )= d d x ( x)+ d y d x or [y 1+x d y d x ]+2 y d y d x = ^ 2 x+ d y d x or (x+2 y-1) d y d x = ^ 2 x-y d y d x = ^ 2 x-y (x+2 y-1)

Differentiation — Previous Year Questions & Solutions