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Class 12 RD Sharma Solutions- Chapter 11 Differentiation – Exercise 11.4 | Set 1

  • Last Updated : 13 Jan, 2021

Find dy/dx in each of the following:

Question 1. xy = c2

Solution: 

We have xy=c2

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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.




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