Class 12 RD Sharma Solutions Chapter 11 Differentiation Exercise 11.4 | Set 2 (original) (raw)
Last Updated : 23 Aug, 2024
**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
**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(√(x 2 +y 2 )) = 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 ((x 2 -y 2 )/(x 2 +y 2 )) = 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(x 2 +1) 1/2 **= log((x 2 +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 = (log cosx sinx)(log sinx cosx) -1 +sin -1 (2x/(1+x 2 ))
**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 = x 2 -y 2
**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. e x +e y **= e x+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 = (cos 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=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
Summary
Exercise 11.4 in RD Sharma's Class 12 Mathematics textbook focuses on the application of differentiation to find the rate of change of quantities with respect to time. This set covers problems involving related rates, where students must determine how quickly one quantity changes in relation to another. The questions typically involve geometric figures or real-world scenarios, requiring students to apply chain rule, implicit differentiation, and other calculus concepts to solve complex rate problems.
Practice Questions:
**1. A stone is dropped into a calm lake, creating circular ripples. If the radius of the ripple increases at a rate of 2 cm/s, how fast is the area of the ripple increasing when the radius is 5 cm?
**2. A ladder 10 meters long rests against a vertical wall. If the bottom of the ladder slides away from the wall at 0.5 m/s, how fast is the top of the ladder sliding down the wall when the bottom is 6 meters from the wall?
**3. A cylindrical tank with radius 3 meters is being filled with water at a rate of 2 m³/min. How fast is the water level rising?
**4. A kite is flying at a height of 100 meters above the ground. If the string is being released at a rate of 8 m/s, at what rate is the kite moving horizontally when 130 meters of string have been released?
**5. A plane is flying horizontally at an altitude of 5 km and a speed of 800 km/h. How fast is the plane's distance from an observer on the ground changing when it is directly over a point 12 km from the observer?
**6. Water is being pumped out of a conical reservoir at a rate of 10 m³/min. The reservoir has a radius of 20 meters and a depth of 12 meters. How fast is the water level decreasing when the water is 9 meters deep?
**7. A balloon is rising at a rate of 5 m/s. A boy is standing 30 meters away from the point directly below the balloon. At what rate is the distance between the boy and the balloon changing when the balloon is 40 meters above the ground?
**8. A particle moves along the curve y = x² + 1. As the particle passes through the point (2, 5), its x-coordinate is increasing at a rate of 3 units per second. How fast is the y-coordinate changing at this instant?
**9. A man walks along a straight path at a speed of 4 km/h. A streetlight is 6 meters above the ground. At what rate is the length of the man's shadow changing when he is 8 meters away from the base of the light?
**10. A sphere is expanding so that its volume increases at a rate of 12π cm³/s. How fast is the radius increasing when the radius is 5 cm?