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

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

^{2}

**Solution: **

We have xy=c

^{2}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.