Class 12 RD Sharma Solutions Chapter 11 Differentiation Exercise 11.4 | Set 1 (original) (raw)
Last Updated : 23 Aug, 2024
**Find dy/dx in each of the following:
**Question 1. xy = c 2
**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 3 **-3xy 2 =x 3 +3x 2 y
**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. x 2/3 +y 2/3 =a 2/3
**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. (x 2 /a 2) + (y 2 /b 2 )=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. x 5 +y 5 =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 (x 2 +y 2 )=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. e x-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-x 2 ) 1/2 +(1-y 2 ) 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-x 2 ) 1/2 +x(1-y 2 ) 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 +y 2 =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 xy 2 =1, prove that 2(dy/dx)+y 3 =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.
Summary
Exercise 11.4 in RD Sharma's Class 12 book focuses on the differentiation of implicit functions. An implicit function is one where the dependent variable is not isolated on one side of the equation. To differentiate implicit functions, we use the technique of implicit differentiation, which involves differentiating both sides of the equation with respect to the independent variable and then solving for the derivative of the dependent variable.