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

  • Last Updated : 13 Jan, 2021
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Find dy/dx in each of the following.

Question 16. If x√(1+y) +y√(1+x) =0  then prove that (1+x)2dy/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



x2(1+y)=y2(1+x)

=>x2+ x2y-y2-y2x=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=0

Hence, proved.

Question 17. log(√(x2+y2)) = tan-1(y/x)

Solution:

We have, 

log(√(x2+y2))=tan-1(y/x)



=>log(x2+y2)(1/2)=tan-1(y/x)

=>(1/2)(log(x2+y2))=tan-1(y/x)

On differentiating both sides with respect to x,

(1/2)d(log(x2+y2))/dx = d(tan-1(y/x))/dx

=>(1/2)*(1/(x2+y2))*(2x+2y(dy/dx))=1/(1+(y/x)2)((x(dy/dx)-y)/x2)

=>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((x2-y2)/(x2+y2)) = a

Solution:

We have,

tan-1((x2-y2)/(x2+y2))=a

=>(x2-y2)/(x2+y2)=tan a

=>(x2-y2)=(tan a)(x2+y2)

On differentiating both sides with respect to x,

=>d(x2-y2)dx=d((tan a)(x2+y2))/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+xy2)/((y+y)x))=-(x+y+x2y)/(y)(x+y))

=>(dy/dx)=-((x+y+x2y)/(x+y+xy2))(y/x)

Therefore, the answer is,

dy/dx=-((x(x2y+x+y))/(y(xy2+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)*cos2(a+y))/(sin(a+y)))+(sina)sin(a+y))=sin(a+y)

(dy/dx)((sina)((cos2(a+y)+sin2(a+y))/sin(a+y)) =sin(a+y)

(dy/dx)=(sin2(a+y))/(sin a)

Therefore, the answer is,

(dy/dx)=sin2(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(x2+1)1/2 = log((x2+1)1/2-x)

Solution:

We have,

y(x2+1)1/2=log((x2+1)1/2-x)

Differentiating it with respect to x,

d(y(x2+1)1/2)/dx=(((x2+1)1/2-x)-1/2)(2(x2+1))-1/2(2x-1)

=>2xy(2(x2+1)-1/2)+(x2+1)1/2(dy/dx)=(((x2+1)1/2-x)-1/2)(x-(x2+1)1/2)(x2+1))-1/2

=>(dy/dx)(x2+1)1/2=((((x2+1)1/2-x)-1/2)(x-(x2+1))-1/2x)/(x2+1))-(xy)(x2+1)-1/2

=>(dy/dx)(x2+1)1/2=(-1/(x2+1)1/2)-(xy)(x2+1)-1/2

=>(dy/dx)(x2+1)1/2=(-1-xy)(x2+1)-1/2

=>(dy/dx)(x2+1)=-(1+xy)

=>(dy/dx)(x2+1)+xy+1=0

Therefore, the answer is,

(dy/dx)(x2+1)+xy+1=0

Question 25. y = (logcosxsinx)(logsinxcosx)-1+sin-1(2x/(1+x2))

Find dy/dx at x=pi/4

Solution:

We have,

y=(logcosxsinx)(logsinxcosx)-1+sin-1(2x/(1+x2))

=>y=(logcosxsinx)(logcosxsinx)+sin-1(2x/(1+x2))

=>y=(logcosxsinx)2+sin-1(2x/(1+x2))

=>y=((log sinx)/log(cosx))2+sin-1(2x/(1+x2))

Differentiating it with respect to x,

dy/dx=d((log sin x)/log(cos x))2/dx+ d(sin-1(2x/(1+x2)))

=>dy/dx =2((log sinx)/log(cosx))d((log sinx)/(log cosx))/dx+1/(√1-((2x)/(1+x2))2d(2x/(1+x2))/dx

=>dy/dx=2((log sinx)/log(cosx))*((log cosx)(1/sin x)*(cos x)-(log sinx)*(1/cosx)*(-sinx))/(log(cosx))2 +((1+x2)/√(1+x4-2x2))((1+x2)2-4x2)/(1+x2)2

=>dy/dx=2(log sinx)/(logcosx)*((log cosx)*cotx+(log sinx)tanx)+((1+x2)/√(1-x2)2)(1-2x2)/(1+x2)2

=>dy/dx=(2(log sinx)*((log cosx)cotx+(log sinx)tanx))/(log cosx)3+2/(1+x2)

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+(pi2)/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 = x2-y2

Solution:

We have,

sin(xy)+y/x=x2-y2

Differentiating it with respect to x,

d(sin(xy)+d(y/x))/dx =d(x2)/dx -d(y2)/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

Solution:

We have,

(y+x)1/2+(y-x)1/2=c

Differentiating 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-((y2-x2)1/2)/(x)

 Therefore, the answer is,

dy/dx=(y-((y2-x2)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

=>sec2(x+y)(d(x+y)/dx)+sec2(x-y)d(x-y)/dx=0

=>sec2(x+y)(1+dy/dx)+sec2(x-y)(1-dy/dx)=0

=>(dy/dx)(sec2(x+y)-sec2(x-y))+sec2(x+y)+sec2(x-y)=0

=>dy/dx=(sec2(x+y)+sec2(x-y))/(sec2(x-y)-sec2(x+y))

Therefore, the answer is,

dy/dx=(sec2(x+y)+sec2(x-y))/(sec2(x-y)-sec2(x+y))

Question 29. ex+ey = ex+y

Solution:



We have,

d(ex+ey)/dx=de(x+y)/dx

=>ex+ey(dy/dx)=e(x+y)(1+(dy/dx))

=>(dy/dx)(ey-e(x+y))=e(x+y)-ex

=>(dy/dx)=(e(x+y)-ex)/(ey-e(x+y))

=>dy/dx=ex(ey-1)/ey(1-ex)

Therefore, the answer is,

dy/dx=ex(ey-1)/ey(1-ex)

Question 30. If cosy = xcos(a+y). Then Prove that, dy/dx = (cos2(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=cos2(a+y)/(cosy*sin(a+y)-siny(cos(a+y)))

=>dy/dx=cos2(a+y)/(sin(a+y-y))

=>dy/dx=cos2(a+y)/sin(a)

Therefore, the answer is,

dy/dx=cos2(a+y)/sin a

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