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

**Find dy/dx in each of the following:**

**Question 1. xy = c**^{2}

^{2}

**Solution: **

We have xy=c

^{2}Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

Demo Class for First Step to Coding Course,specificallydesigned for students of class 8 to 12.The students will get to learn more about the world of programming in these

free classeswhich will definitely help them in making a wise career choice in the future.Differentiating both sides with respect to x.

d(xy)/dx = d(c

^{2})/dxBy 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

^{3 }-3xy

^{2}=x

^{3}+3x

^{2}y

**Solution:**

We have

y

^{3}-3xy^{2}=x^{3}+3x^{2}yDifferentiating both sides with respect to x,

d(y

^{3}-3xy^{2})/dx=d(x^{3}+3x^{2}y)/dxBy product rule,

=>

3y^{ }^{2}dy/dx-3y^{2}-6xydy/dx=3x^{2}+3x^{2}dy/dx+6xy=>3y

^{2}dy/dx-6xydy/dx-3x^{2}dy/dx=3x^{2}+3y^{2}+6xy=>dy/dx(3y

^{2}-3x^{2}-6xy)=3x^{2}+3y^{2}+6xy=>3dy/dx(y

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

^{2}}/{3(y^{2}-x^{2}-2xy)Therefore the answer is,

dy/dx=(x+y)

^{2}/(y^{2}-x^{2}-2xy)

**Question 3. x**^{2/3}+y^{2/3}=a^{2/3}

^{2/3}+y

^{2/3}=a

^{2/3}

**Solution:**

We have,

x

^{2/3}+y^{2/3}=a^{2/3}Differentiating both sides with respect to x,

d(x

^{2/3})/dx +d(y^{2/3})/dx=d(a^{2/3})/dx=> 2/3x

^{1/3 }+(2/3y^{1/3})dy/dx=0=>1/x

^{1/3 }+(1/y^{1/3})dy/dx =0=> dy/dx=-y

^{1/3}/x^{1/3}Therefore the answer is,

dy/dx=-y

^{1/3}/x^{1/3}

**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

^{2}/a

^{2)}+ (y

^{2}/b

^{2})=1

**Solution:**

We have,

(x

^{2}/a^{2})+(y^{2}/b^{2})=1Differentiating both sides with respect to x,

d(x

^{2}/a^{2})/dx +d(y^{2}/b^{2})/dx =d(1)/dx=>(2x/a

^{2})+(2y/b^{2})(dy/dx)=0=>(y/b

^{2})(dy/dx)=-x/a^{2}=>dy/dx = -xb

^{2}/ya^{2}Therefore the answer is,

dy/dx =-xb

^{2}/ya^{2}.

**Question 6. x**^{5}+y^{5}=5xy

^{5}+y

^{5}=5xy

**Solution:**

We have,

x

^{5}+y^{5 }=5xyDifferentiating both sides with respect to x,

d(x

^{5})/dx +d(y^{5})/dx=d(5xy)/ dx=> 5x

^{4}+ 5y^{4}dy/dx=5y+ (5x)dy/dx=>y

^{4(}dy/dx)-x(dy/dx)=y-x^{4}=>dy/dx(y

^{4}-x)=y-x^{4}=>dy/dx=(y-x

^{4})/(y^{4}-x)Therefore the answer is,

dy/dx=(y-x

^{4})/(y^{4}-x)

**Question 7. (x+y)**^{2}=2axy

^{2}=2axy

**Solution:**

We have,

(x+y)

^{2}=2axyDifferentiating 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. (x^{2}+y^{2})^{2}=xy

**Solution:**

We have,

(x

^{2}+y^{2})^{2}=xyDifferentiating both sides with respect to x,

d(x

^{2}+y^{2})^{2}/dx=d(xy)/dx=>2(x

^{2}+y^{2})(2x+2y(dy/dx))=y+x(dy/dx)=>4x(x

^{2}+y^{2})+4y(x^{2}+y^{2})(dy/dx)=y+x(dy/dx)=>4y(x

^{2}+y^{2})(dy/dx)-x(dy/dx)=y-4x(x^{2}+y^{2})=>(dy/dx)(4y(x

^{2}+y^{2})-x)=y-4x(x^{2}+y^{2})=>dy/dx=(y-4x(x

^{2}+y^{2}))/(4y(x^{2}+y^{2})-x)Therefore the answer is,

dy/dx=(y-4x(x

^{2}+y^{2}))/(4y(x^{2}+y^{2})-x)

**Question 9. tan**^{-1}(x^{2}+y^{2})=a

^{-1}(x

^{2}+y

^{2})=a

**Solution:**

We have,

tan

^{-1}(x^{2}+y^{2})=aDifferentiating both sides with respect to x ,

d(tan

^{-1}(x^{2}+y^{2}))/dx=da/dx=>(1/(x

^{2}+y^{2}))(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)

^{x-y}=log(x/y)

**Solution:**

We have,

e

^{x-y}=log(x/y)=>e

^{x-y}=log x -log yDifferentiating both sides with respect to x,

d(e

^{x-y})/dx=d(log x- log y)/dx=>e

^{x-y}(1-dy/dx)=1/x-(1/y)(dy/dx)=>e

^{x-y}-e^{x-y}(dy/dx)=1/x -(1/y)(dy/dx)=>(1/y)(dy/dx) – e

^{x-y}(dy/dx)=1/x-e^{x-y}=> dy/dx((1/y)-e

^{x-y})=(1-xe^{x-y})/x=> (dy/dx)(1-ye

^{x-y})/y=(1-xe^{x-y})/x=>dy/dx=y(1-xe

^{x-y})/x(1-ye^{x-y})Therefore the answer is,

dy/dx=y(1-xe

^{x-y})/x(1-ye^{x-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)

^{2})

^{1/2}+(1-y

^{2})

^{1/2}=a(x-y)

**Solution:**

We have,

(1-x

^{2})^{1/2}+(1-y^{2})^{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

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

^{-1}a=((A-B)/2)Differentiating both sides with respect to x,

d(2cot

^{-1}a)/dx=d(A-B)/dx=>0=d(sin

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

^{2})^{1/2}) -(1/(1-y^{2})^{1/2})*(dy/dx)=>(1/(1-y

^{2})^{1/2})*dy/dx=1/((1-x^{2})^{1/2})=>dy/dx=((1-y

^{2})^{1/2})/(1-x^{2})^{1/2}Therefore, the answer is,

dy/dx=((1-y

^{2})^{1/2})/(1-x^{2})^{1/2}

**Question 13. y(1-x**^{2})^{1/2}+x(1-y^{2})^{1/2}=1

^{2})

^{1/2}+x(1-y

^{2})

^{1/2}=1

**Solution:**

We have,

y(1-x

^{2})^{1/2}+x(1-y^{2})^{1/2}=1Let, 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

^{-1}x +sin^{-1}y=22/14Differentiating both sides with respect to x,

d(sin

^{-1}x)/dx +d(sin^{-1 }y)/dx=d(22/14)/dx=>1/((1-x

^{2})^{1/2})+ (1/((1-y^{2})^{1/2}))(dy/dx)=0=>dy/dx=-((1-y

^{2})^{1/2})/((1-x^{2})^{1/2})Therefore, the answer is,

dy/dx=-((1-y

^{2})^{1/2})/((1-x^{2})^{1/2})

**Question 14. If xy=1, prove that dy/dx +y**^{2}=0

^{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+y

^{2}=0Hence, proved.

**Question 15. If xy**^{2}=1, prove that 2(dy/dx)+y^{3}=0

^{2}=1, prove that 2(dy/dx)+y

^{3}=0

**Solution:**

We have,

xy

^{2}=1Differentiating with respect to x,

d(xy

^{2})/dx=d1/dx=>2xy(dy/dx)+y

^{2}=0=>dy/dx=-y

^{2}/2xy=>dy/dx =-y/2x

Also x=1/y

^{2}So, dy/dx=-y(y

^{2})/2=>2dy/dx=-y

^{3}2dy/d+y

^{3}=0Hence, proved.