# Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.4 | Set 2

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

**Question 16. If x√(1+y) +y√(1+x) =0 then prove that (1+x)**^{2}dy/dx +1=0

^{2}dy/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

x

^{2}(1+y)=y^{2}(1+x)=>x

^{2}+ x^{2}y-y^{2}-y^{2}x=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=0Hence, proved.

**Question 17. log(√(x**^{2}+y^{2})) = tan^{-1}(y/x)

^{2}+y

^{2})) = tan

^{-1}(y/x)

**Solution:**

We have,

log(√(x

^{2}+y^{2}))=tan^{-1}(y/x)=>log(x

^{2}+y^{2})^{(1/2)}=tan^{-1}(y/x)=>(1/2)(log(x

^{2}+y^{2}))=tan^{-1}(y/x)On differentiating both sides with respect to x,

(1/2)d(log(x

^{2}+y^{2}))/dx = d(tan^{-1}(y/x))/dx=>(1/2)*(1/(x

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

^{-1}((x

^{2}-y

^{2})/(x

^{2}+y

^{2})) = a

**Solution:**

We have,

tan

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

^{2}-y^{2})/(x^{2}+y^{2})=tan a=>(x

^{2}-y^{2})=(tan a)(x^{2}+y^{2})On differentiating both sides with respect to x,

=>d(x

^{2}-y^{2})dx=d((tan a)(x^{2}+y^{2}))/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+xy

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

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

dy/dx=-((x(x

^{2}y+x+y))/(y(xy^{2}+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)*cos

^{2}(a+y))/(sin(a+y)))+(sina)sin(a+y))=sin(a+y)(dy/dx)((sina)((cos

^{2}(a+y)+sin^{2}(a+y))/sin(a+y)) =sin(a+y)(dy/dx)=(sin

^{2}(a+y))/(sin a)Therefore, the answer is,

(dy/dx)=sin

^{2}(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(x**^{2}+1)^{1/2 }= log((x^{2}+1)^{1/2}-x)

^{2}+1)

^{1/2 }= log((x

^{2}+1)

^{1/2}-x)

**Solution:**

We have,

y(x

^{2}+1)^{1/2}=log((x^{2}+1)^{1/2}-x)Differentiating it with respect to x,

d(y(x

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

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

^{2}+1)^{1/2}=((((x^{2}+1)^{1/2}-x)^{-1/2})(x-(x^{2}+1))^{-1/2}x)/(x^{2}+1))-(xy)(x^{2}+1)^{-1/2}=>(dy/dx)(x

^{2}+1)^{1/2}=(-1/(x^{2}+1)^{1/2})-(xy)(x^{2}+1)^{-1/2}=>(dy/dx)(x

^{2}+1)^{1/2}=(-1-xy)(x^{2}+1)^{-1/2}=>(dy/dx)(x

^{2}+1)=-(1+xy)=>(dy/dx)(x

^{2}+1)+xy+1=0Therefore, the answer is,

(dy/dx)(x

^{2}+1)+xy+1=0

**Question 25. y = (log**_{cosx}sinx)(log_{sinx}cosx)^{-1}+sin^{-1}(2x/(1+x^{2}))

_{cosx}sinx)(log

_{sinx}cosx)

^{-1}+sin

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

^{2}))

**Find dy/dx at x=pi/4**

**Solution:**

We have,

y=(log

_{cosx}sinx)(log_{sinx}cosx)^{-1}+sin^{-1}(2x/(1+x^{2}))=>y=(log

_{cosx}sinx)(log_{cosx}sinx)+sin^{-1}(2x/(1+x^{2}))=>y=(log

_{cosx}sinx)^{2}+sin^{-1}(2x/(1+x^{2}))=>y=((log sinx)/log(cosx))

^{2}+sin^{-1}(2x/(1+x^{2}))Differentiating it with respect to x,

dy/dx=d((log sin x)/log(cos x))

^{2}/dx+ d(sin^{-1}(2x/(1+x^{2})))=>dy/dx =2((log sinx)/log(cosx))d((log sinx)/(log cosx))/dx+1/(√1-((2x)/(1+x

^{2}))^{2}d(2x/(1+x^{2}))/dx=>dy/dx=2((log sinx)/log(cosx))*((log cosx)(1/sin x)*(cos x)-(log sinx)*(1/cosx)*(-sinx))/(log(cosx))

^{2}+((1+x^{2})/√(1+x^{4}-2x^{2}))((1+x^{2})2-4x^{2})/(1+x^{2})^{2}=>dy/dx=2(log sinx)/(logcosx)*((log cosx)*cotx+(log sinx)tanx)+((1+x

^{2})/√(1-x^{2})^{2})(1-2x^{2})/(1+x^{2})^{2}=>dy/dx=(2(log sinx)*((log cosx)cotx+(log sinx)tanx))/(log cosx)

^{3}+2/(1+x^{2})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+(pi^{2})/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 = x**^{2}-y^{2}

^{2}-y

^{2}

**Solution:**

We have,

sin(xy)+y/x=x

^{2}-y^{2}Differentiating it with respect to x,

d(sin(xy)+d(y/x))/dx =d(x

^{2})/dx -d(y^{2})/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

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

^{1/2 }= c

**Solution:**

We have,

(y+x)

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

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

dy/dx=(y-((y

^{2}-x^{2})^{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

=>sec

^{2}(x+y)(d(x+y)/dx)+sec^{2}(x-y)d(x-y)/dx=0=>sec

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

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

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

dy/dx=(sec

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

**Question 29. e**^{x}+e^{y }= e^{x+y}

^{x}+e

^{y }= e

^{x+y}

**Solution:**

We have,

d(e

^{x}+e^{y})/dx=de^{(x+y)}/dx=>e

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

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

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

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

dy/dx=e

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

**Question 30. If cosy = xcos(a+y). Then Prove that, dy/dx = (cos**^{2}(a+y))/sin a

^{2}(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=cos

^{2}(a+y)/(cosy*sin(a+y)-siny(cos(a+y)))=>dy/dx=cos

^{2}(a+y)/(sin(a+y-y))=>dy/dx=cos

^{2}(a+y)/sin(a)Therefore, the answer is,

dy/dx=cos

^{2}(a+y)/sin a

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