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.
| Textbook | NCERT |
| Class | Class 12th |
| Subject | Maths |
| Chapter | 11 |
| Exercise | 11.4 |
| Category | RD 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. 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)
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+y5 =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+ey = 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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