# RD Sharma Class 12 Ex 11.4 Solutions Chapter 11 Differentiation

Here we provide RD Sharma Class 12 Ex 11.4 Solutions Chapter 11 Differentiation for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 11.4 Solutions Chapter 11 Differentiation book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 11.4 Solutions Chapter 11 Differentiation

### Question 1. xy = c2

Solution:

We have xy=c2

Differentiating both sides with respect to x.

d(xy)/dx = d(c2)/dx

By product rule,

y+x*dy/dx=0

dy/dx=-y/x

### Question 2. y3 -3xy2=x3+3x2y

Solution:

We have

y3-3xy2=x3+3x2y

Differentiating both sides with respect to x,

d(y3-3xy2)/dx=d(x3+3x2y)/dx

By product rule,

=> 3y2dy/dx-3y2-6xydy/dx=3x2+3x2dy/dx+6xy

=>3y2dy/dx-6xydy/dx-3x2dy/dx=3x2+3y2+6xy

=>dy/dx(3y2-3x2-6xy)=3x2+3y2+6xy

=>3dy/dx(y2-x2-2xy)=3(x2+y2+2xy)

=> dy/dx={3(x+y)2}/{3(y2-x2-2xy)

dy/dx=(x+y)2/(y2-x2-2xy)

### Question 3. x2/3+y2/3=a2/3

Solution:

We have,

x2/3+y2/3=a2/3

Differentiating both sides with respect to x,

d(x2/3)/dx +d(y2/3)/dx=d(a2/3)/dx

=> 2/3x1/3 +(2/3y1/3)dy/dx=0

=>1/x1/3 +(1/y1/3)dy/dx =0

=> dy/dx=-y1/3/x1/3

dy/dx=-y1/3/x1/3

### Question 4. 4x+3y=log(4x-3y)

Solution:

We have,

4x+3y= log(4x-3y)

Differentiating both sides with respect to x,

d(4x+3y)/dx=d(log(4x-3y))/dx

=>4+3dy/dx=(1/(4x-3y))(4-3dy/dx)

=>3dy/dx+3dy/dx(1/4x-3y)=4/(4x-3y)-4

=>(3dy/dx)(1+1/(4x-3y))=(4-16x+12y)/(4x-3y)

=>(3dy/dx)((4x-3y+1)/(4x-3y))=(4-16x+12y)/(4x-3y)

=>(3dy/dx)(4x-3y+1)=4-16x+12y

=>(3dy/4dx)(4x-3y+1)=3y-4x+1

=>dy/dx=(4/3)((3y-4x+1)/(4x-3y+1))

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

### Question 5. (x2/a2)+ (y2/b2)=1

Solution:

We have,

(x2/a2)+(y2/b2)=1

Differentiating both sides with respect to x,

d(x2/a2)/dx +d(y2/b2)/dx =d(1)/dx

=>(2x/a2)+(2y/b2)(dy/dx)=0

=>(y/b2)(dy/dx)=-x/a2

=>dy/dx = -xb2/ya2

dy/dx =-xb2/ya2.

### Question 6. x5+y5=5xy

Solution:

We have,

x5+y=5xy

Differentiating both sides with respect to x,

d(x5)/dx +d(y5)/dx=d(5xy)/ dx

=> 5x4 + 5y4dy/dx=5y+ (5x)dy/dx

=>y4(dy/dx)-x(dy/dx)=y-x4

=>dy/dx(y4-x)=y-x4

=>dy/dx=(y-x4)/(y4-x)

dy/dx=(y-x4)/(y4-x)

### Question 7. (x+y)2=2axy

Solution:

We have,

(x+y)2=2axy

Differentiating with respect to x,

d(x+y)2/dx=d(2axy)/dx

=>2(x+y)(1+dy/dx)=2ax(dy/dx) +2ay

=>(x+y)+(x+y)dy/dx =ax(dy/dx)+ay

=>(x+y)dy/dx-ax(dy/dx)=ay-x-y()

=>(dy/dx)(x+y-ax)=ay-x-y

=>dy/dx=(ay-x-y)/(x+y-ax)

dy/dx=(ay-x-y)/(x+y-ax)

### Question 8. (x2+y2)2=xy

Solution:

We have,

(x2+y2)2=xy

Differentiating both sides with respect to x,

d(x2+y2)2/dx=d(xy)/dx

=>2(x2+y2)(2x+2y(dy/dx))=y+x(dy/dx)

=>4x(x2+y2)+4y(x2+y2)(dy/dx)=y+x(dy/dx)

=>4y(x2+y2)(dy/dx)-x(dy/dx)=y-4x(x2+y2)

=>(dy/dx)(4y(x2+y2)-x)=y-4x(x2+y2)

=>dy/dx=(y-4x(x2+y2))/(4y(x2+y2)-x)

dy/dx=(y-4x(x2+y2))/(4y(x2+y2)-x)

### Question 9. tan-1(x2+y2)=a

Solution:

We have,

tan-1(x2+y2)=a

Differentiating both sides with respect to x ,

d(tan-1(x2+y2))/dx=da/dx

=>(1/(x2+y2))(2x+2y(dy/dx))=0

=>x+y(dy/dx)=0

=> dy/dx=-x/y

dy/dx=-x/y

### Question 10. ex-y=log(x/y)

Solution:

We have,

ex-y=log(x/y)

=>ex-y=log x -log y

Differentiating both sides with respect to x,

d(ex-y)/dx=d(log x- log y)/dx

=>ex-y(1-dy/dx)=1/x-(1/y)(dy/dx)

=>ex-y -ex-y(dy/dx)=1/x -(1/y)(dy/dx)

=>(1/y)(dy/dx) – ex-y(dy/dx)=1/x-ex-y

=> dy/dx((1/y)-ex-y)=(1-xex-y)/x

=> (dy/dx)(1-yex-y)/y=(1-xex-y)/x

=>dy/dx=y(1-xex-y)/x(1-yex-y)

dy/dx=y(1-xex-y)/x(1-yex-y)

### Question 11. sin(xy)+ cos(x+y)=1

Solution:

We have,

sin(xy)+ cos(x+y)=1

Differentiating both sides with respect to x,

d(sin(xy))/dx + d(cos(x+y))/dx=d1/dx

=>cos(xy)(y+xdy/dx) +(-sin(x+y)(1+dy/dx)= 0

=>cos(xy)(y+xdy/dx) = (sin(x+y)(1+dy/dx)

=>ycos(xy)+x*cos(xy)*(dy/dx)= sin(x+y) + sin(x+y)* (dy/dx)

=>x*cos(xy)*(dy/dx) – sin(x+y)* (dy/dx) = sin(x+y) – ycos(xy)

=>(dy/dx)((x*cos(xy))-sin(x+y))= sin(x+y) – ycos(xy)

=>dy/dx =(sin(x+y)-ycos(xy))/((x*cos(xy))-sin(x+y))

dy/dx=(sin(x+y)-ycos(xy))/((x*cos(xy))-sin(x+y))

### Question 12. (1-x2)1/2+(1-y2)1/2=a(x-y)

Solution:

We have,

(1-x2)1/2+(1-y2)1/2=a(x-y)

Let x=sin A and y= sin B

So the expression becomes,

cosA + cosB=a(sinA-sinB)

=>a=(cosA+cosB)/(sinA-sinB)

=>a=(2(cos((A+B)/2))*(cos((A-B)/2)))/(2cos((A+B)/2)*sin((A-B)/2)))

=> a =(cos(A-B)/2)/(sin(A-B)/2)

=> a=cot((A-B)/2)

=>cot-1a=((A-B)/2)

=>2cot-1a=((A-B)/2)

Differentiating both sides with respect to x,

d(2cot-1a)/dx=d(A-B)/dx

=>0=d(sin-1x)/dx -d(sin-1y)/dx

=> 0 = 1/((1-x2)1/2) -(1/(1-y2)1/2)*(dy/dx)

=>(1/(1-y2)1/2)*dy/dx=1/((1-x2)1/2)

=>dy/dx=((1-y2)1/2)/(1-x2)1/2

dy/dx=((1-y2)1/2)/(1-x2)1/2

### Question 13. y(1-x2)1/2+x(1-y2)1/2=1

Solution:

We have,

y(1-x2)1/2+x(1-y2)1/2=1

Let, x=sin A and y=sin B

So, the expression becomes,

(sin B)*(cos A)+(sin A)*(cos B) =1

=> sin(A+B) =1

=> sin-1(1) =A+B

=>A+B =22/(7*2)

=>sin-1x +sin-1y=22/14

Differentiating both sides with respect to x,

d(sin-1x)/dx +d(sin-1 y)/dx=d(22/14)/dx

=>1/((1-x2)1/2)+ (1/((1-y2)1/2))(dy/dx)=0

=>dy/dx=-((1-y2)1/2)/((1-x2)1/2)

dy/dx=-((1-y2)1/2)/((1-x2)1/2)

### Question 14. If xy=1, prove that dy/dx +y2=0

Solution:

We have,

xy=1

Differentiating both sides with respect to x,

d(xy)/dx =d1/dx

=>x(dy/dx)+y=0

=>dy/dx =-y/x

Also x=1/y

so,   dy/dx=-y(y)

=>dy/dx+y2=0

Hence, proved.

### Question 15. If xy2=1, prove that 2(dy/dx)+y3=0

Solution:

We have,

xy2=1

Differentiating with respect to x,

d(xy2)/dx=d1/dx

=>2xy(dy/dx)+y2 =0

=>dy/dx=-y2/2xy

=>dy/dx =-y/2x

Also x=1/y2

So, dy/dx=-y(y2)/2

=>2dy/dx=-y3

2dy/d+y3=0

Hence, proved.

Find dy/dx in each of the following.
Question 16. If x√(1+y) +y√(1+x) =0  then prove that (1+x)2dy/dx +1=0
Solution:

We have,
x√(1+y) +y√(1+x) =0
=>x√(1+y)=-y√(1+x)
On squaring both sides, we have
x2(1+y)=y2(1+x)
=>x2+ x2y-y2-y2x=0
=>(x+y)(x-y)+xy(x-y)=0
=>(x-y)(x+y+xy)=0
So, either (x-y)=0
or, x+y+xy=0
=>x+y(1+x)=0
=>y=-x/(1+x)
On differentiating both sides with respect to x,
dy/dx=d(-x/(1+x))/dx
On applying quotient rule,
dy/dx = ((x)1-(1+x))/(1+x)2
=>dy/dx=(-1/(1+x)2)
=>(dy/dx)(1+x)2+1=0
Hence, proved.
Question 17. log(√(x2+y2)) = tan-1(y/x)
Solution:
We have,
log(√(x2+y2))=tan-1(y/x)
=>log(x2+y2)(1/2)=tan-1(y/x)
=>(1/2)(log(x2+y2))=tan-1(y/x)
On differentiating both sides with respect to x,
(1/2)d(log(x2+y2))/dx = d(tan-1(y/x))/dx
=>(1/2)*(1/(x2+y2))*(2x+2y(dy/dx))=1/(1+(y/x)2)((x(dy/dx)-y)/x2)
=>x+y(dy/dx)=x(dy/dx)-y
=>(dy/dx)(y-x)=-(x+y)
=>(dy/dx)=(x+y/(x-y)
(dy/dx)=(x+y)/(x-y)
Question 18. sec((x+y)/(x-y)) = a
Solution:
We have,
sec((x+y)/(x-y))=a
=>(x+y)/(x-y)=sec-1(a)
On differentiating both sides with respect to x,
=>d((x+y)/(x-y))/dx=d(sec-1(a))/dx
=>(x-y)(1+(dy/dx))-(x+y)(1-(dy/dx))=0(x-y)2
=>x-y+(x-y)(dy/dx)-(x+y)+(x+y)(dy/dx)=0
=>(dy/dx)(x-y+x+y)-2y=0
=>(dy/dx)(2x)=2y
=>(dy/dx)=(y/x)
(dy/dx)=(y/x)
Question 19. tan-1((x2-y2)/(x2+y2)) = a
Solution:
We have,
tan-1((x2-y2)/(x2+y2))=a
=>(x2-y2)/(x2+y2)=tan a
=>(x2-y2)=(tan a)(x2+y2)
On differentiating both sides with respect to x,
=>d(x2-y2)dx=d((tan a)(x2+y2))/dx
=>2x-2y(dy/dx)=(tan a)(2x+2y(dy/dx))
=>x-y(dy/dx)=(tan a)(x+y(dy/dx))
=>-(dy/dx)(y+y(tan a))=x(tan a)-x
=>-(dy/dx)=(x(tan a-1))/(y(1+tan a))
=>dy/dx= (x(1-tan a))/(y(1+tan a))
dy/dx =(x(1-tana))/(y(1+tan a))
Question 20. xy(log(x+y)) = 1
Solution:
We have,
xy(log(x+y))=1
Differentiating it with respect to x,
d(xy(log(x+y)))/dx =d1/dx
=>y(log(x+y))+x(log(x+y)dy/dx+((xy)/(x+y))(1+(dy/dx)))=0
=>y(log(x+y))+((xy)/(x+y))+(dy/dx)(x(log(x+y))+(xy)/(x+y))=0
=>(dy/dx)(x(log(x+y))+(xy)/(x+y))=-(y(log(x+y))+(xy)/(x+y))
It can be deduced that ,
y(log(x+y))=1/x
x(log(x+y))=1/y
So,
(dy/dx)((1/y)+(xy)/(x+y))=-((1/x)+(xy)/(x+y)
=>(dy/dx)((x+y+xy2)/((y+y)x))=-(x+y+x2y)/(y)(x+y))
=>(dy/dx)=-((x+y+x2y)/(x+y+xy2))(y/x)
dy/dx=-((x(x2y+x+y))/(y(xy2+x+y)))
Question 21. y = xsin(a+y)
Solution:
We have,
y=x sin(a+y)
Differentiating it with respect to x,
dy/dx=sin(a+y) +x*cos(a+y){0+dy/dx}
=>dy/dx =sin(a+y) +x*cos(a+y)*(dy/dx)
=>dy/dx-x*cos(a+y)*(dy/dx) =sin(a+y)
=>(dy/dx)(1-x*cos(a+y))=sin(a+y)
=>dy/dx=(sin(a+y))/(1-x*cos(a+y))
dy/dx =(sin(a+y))/(1-x*cos(a+y))
Question 22. x*sin(a+y)+(sin a)*(cos(a+y)) = 0
Solution:
We have,
x*sin(a+y)+(sin a)*(cos(a+y))=0
On differentiating both sides with respect to x,
d(x*sin(a+y)+(sin a)*(cos(a+y)))/dx=d0/dx
=>sin(a+y)+x*cos(a+y)*(dy/dx)-(sin a)sin(a+y)(dy/dx)=0
=>(dy/dx)(xcos(a+y)-sina(sin(a+y)))=-sin(a+y)
=>(dy/dx)=sin(a+y)/(sina*sin(a+y)-xcos(a+y))
From above,
x=-((sina)*cos(a+y))/sin(a+y)
Putting in the above equation,
(dy/dx)*(((sina)*cos2(a+y))/(sin(a+y)))+(sina)sin(a+y))=sin(a+y)
(dy/dx)((sina)((cos2(a+y)+sin2(a+y))/sin(a+y)) =sin(a+y)
(dy/dx)=(sin2(a+y))/(sin a)
(dy/dx)=sin2(a+y)/(sina)
Question 23. y = x*siny
Solution:
We have,
y=x*siny
On differentiating both sides with respect to x,
dy/dx=siny+x(cosy)(dy/dx)
=>dy/dx-x(cosy)(dy/dx)=siny
=>(dy/dx)(1-x(cosy))=siny
=>dy/dx=(siny)/(1-x(cosy))
(dy/dx)=(siny)/(1-x(cosy))
Question 24. y(x2+1)1/2 = log((x2+1)1/2-x)
Solution:
We have,
y(x2+1)1/2=log((x2+1)1/2-x)
Differentiating it with respect to x,
d(y(x2+1)1/2)/dx=(((x2+1)1/2-x)-1/2)(2(x2+1))-1/2(2x-1)
=>2xy(2(x2+1)-1/2)+(x2+1)1/2(dy/dx)=(((x2+1)1/2-x)-1/2)(x-(x2+1)1/2)(x2+1))-1/2
=>(dy/dx)(x2+1)1/2=((((x2+1)1/2-x)-1/2)(x-(x2+1))-1/2x)/(x2+1))-(xy)(x2+1)-1/2
=>(dy/dx)(x2+1)1/2=(-1/(x2+1)1/2)-(xy)(x2+1)-1/2
=>(dy/dx)(x2+1)1/2=(-1-xy)(x2+1)-1/2
=>(dy/dx)(x2+1)=-(1+xy)
=>(dy/dx)(x2+1)+xy+1=0
(dy/dx)(x2+1)+xy+1=0
Question 25. y = (logcosxsinx)(logsinxcosx)-1+sin-1(2x/(1+x2))
Find dy/dx at x=pi/4
Solution:
We have,
y=(logcosxsinx)(logsinxcosx)-1+sin-1(2x/(1+x2))
=>y=(logcosxsinx)(logcosxsinx)+sin-1(2x/(1+x2))
=>y=(logcosxsinx)2+sin-1(2x/(1+x2))
=>y=((log sinx)/log(cosx))2+sin-1(2x/(1+x2))
Differentiating it with respect to x,
dy/dx=d((log sin x)/log(cos x))2/dx+ d(sin-1(2x/(1+x2)))
=>dy/dx =2((log sinx)/log(cosx))d((log sinx)/(log cosx))/dx+1/(√1-((2x)/(1+x2))2d(2x/(1+x2))/dx
=>dy/dx=2((log sinx)/log(cosx))*((log cosx)(1/sin x)*(cos x)-(log sinx)*(1/cosx)*(-sinx))/(log(cosx))2 +((1+x2)/√(1+x4-2x2))((1+x2)2-4x2)/(1+x2)2
=>dy/dx=2(log sinx)/(logcosx)*((log cosx)*cotx+(log sinx)tanx)+((1+x2)/√(1-x2)2)(1-2x2)/(1+x2)2
=>dy/dx=(2(log sinx)*((log cosx)cotx+(log sinx)tanx))/(log cosx)3+2/(1+x2)
At x=pi/4
dy/dx=(2(log sin(pi/4))*((log(cos(pi/4) cot(pi/4)+(log sin(pi/4))tan(pi/4))/(log cos(pi/4))3+2/(1+(pi2)/16)
=>dy/dx=2(log(1/√2))*(log(1/√2)+log(1/√2))/(log(1/√2))3+32/(16+(pi)2)
=>dy/dx=4(1/(log(1/√2)))+32/(16+(pi)2)
=>dy/dx=4(1/((-1/2)log2)+32/(16+(pi)2)
=>dy/dx=32/(16+(pi)2)-8(1/log2)
(dy/dx)=32/(16+(pi)2)-8(1/log2)
Question 26. sin(xy)+y/x = x2-y2
Solution:
We have,
sin(xy)+y/x=x2-y2
Differentiating it with respect to x,
d(sin(xy)+d(y/x))/dx =d(x2)/dx -d(y2)/dx
=>cos(xy)(x(dy/dx)+y) +(x(dy/dx)-y)(x-2)=2x-2y(dy/dx)
=>x*cos(xy)(dy/dx) + ycos(xy)+(x-1)(dy/dx)-y(x-2)+2y(dy/dx)=2x
=>(dy/dx)(x*cos(xy)+x-1+2y)=2x-ycos(xy)-y(x-2)
=>dy/dx=(2x-ycos(xy)-y(x-2))/(x*cos(xy)+x-1+2y)
(dy/dx)=(2x-ycos(xy)-y(x-2))/(x*cos(xy)+x-1+2y)
Question 27. (y+x)1/2+(y-x)1/2 = c
Solution:
We have,
(y+x)1/2+(y-x)1/2=c
Differentiating it with respect to x,
(1/2)(y+x)-1/2((dy/dx)+1) + (1/2)(y-x)-1/2((dy/dx)-1)=0
=>(dy/dx)((1/2)(y+x)-1/2+(1/2)(y-x)-1/2) +(1/2)(y+x)-1/2-(1/2)(y-x)-1/2=0
=>(dy/dx)=((y-x)-1/2-(y+x)-1/2)/((y+x)-1/2+(y-x)-1/2)
By rationalisation of denominator,
(dy/dx)=(y+x)+(y-x)-2(y+x)1/2(y-x)1/2
=>dy/dx=(2y-2(y+x)1/2(y-x)1/2)/(x+y-y+x)
=>dy/dx=(y-((y2-x2)1/2)/(x)
dy/dx=(y-((y2-x2)1/2)/(x)
Question 28. tan(x+y)+tan(x-y) = 1
Solution:
We have,
tan(x+y)+tan(x-y)=1
Differentiating it with respect to x,
d(tan(x+y)+tan(x-y))/dx=d1/dx
=>sec2(x+y)(d(x+y)/dx)+sec2(x-y)d(x-y)/dx=0
=>sec2(x+y)(1+dy/dx)+sec2(x-y)(1-dy/dx)=0
=>(dy/dx)(sec2(x+y)-sec2(x-y))+sec2(x+y)+sec2(x-y)=0
=>dy/dx=(sec2(x+y)+sec2(x-y))/(sec2(x-y)-sec2(x+y))
dy/dx=(sec2(x+y)+sec2(x-y))/(sec2(x-y)-sec2(x+y))
Question 29. ex+e= ex+y
Solution:
We have,
d(ex+ey)/dx=de(x+y)/dx
=>ex+ey(dy/dx)=e(x+y)(1+(dy/dx))
=>(dy/dx)(ey-e(x+y))=e(x+y)-ex
=>(dy/dx)=(e(x+y)-ex)/(ey-e(x+y))
=>dy/dx=ex(ey-1)/ey(1-ex)
dy/dx=ex(ey-1)/ey(1-ex)
Question 30. If cosy = xcos(a+y). Then Prove that, dy/dx = (cos2(a+y))/sin a
Solution:
We have,
cosy=x*cos(a+y)
Differentiating it with respect to x,
d(cosy)/dx=d(x*cos(a+y))/dx
=>-siny(dy/dx)=cos(a+y)-xsin(a+y)(dy/dx)
=>xsin(a+y)(dy/dx)-siny(dy/dx)=cos(a+y)
=>(dy/dx)(xsin(a+y)-siny)=cos(a+y)
=>dy/dx=(cos(a+y))/(x*sin(a+y)-siny)
Also, x=cosy/cos(a+y)
Substituting it in the earlier statement,
(dy/dx)=(cos(a+y))/((cosy)sin(a+y)/cos(a+y))-siny)
=>dy/dx=cos2(a+y)/(cosy*sin(a+y)-siny(cos(a+y)))
=>dy/dx=cos2(a+y)/(sin(a+y-y))
=>dy/dx=cos2(a+y)/sin(a)