RD Sharma Class 12 Ex 11.4 Solutions Chapter 11 Differentiation

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter11
Exercise11.4
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 11.4 Solutions Chapter 11 Differentiation

Find dy/dx in each of the following:

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

Therefore the answer is.

 dy/dx=-y/x

Question 2. y-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)

Therefore the answer is,

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

Therefore the answer is,

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))

Therefore the answer is,

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

Therefore the answer is,

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)

Therefore the answer is,

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)

Therefore  the answer is,

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)

Therefore the answer is,

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

Therefore the answer is,

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)

Therefore the answer is,

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))

Therefore, the answer is,

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

Therefore, the answer is,

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)

Therefore, the answer is,

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)
Therefore, the answer is,
(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)
Therefore, the answer is,
(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))
Therefore, the answer is,
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)
Therefore, the answer is,
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))
Therefore, the answer is,
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)
Therefore, the answer is,
(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))
Therefore, the answer is,
(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
Therefore, the answer is,
(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)
 Therefore, the answer is,
(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)
 Therefore, the answer is,
(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)
 Therefore, the answer is,
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))
Therefore, the answer is,
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)
Therefore, the answer is,
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)
Therefore, the answer is,
dy/dx=cos2(a+y)/sin a

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