Continuity & Differentiability NCERT Solutions
NCERT Exercise 5.3
Find `dy/dx` in the following:
Question: 1 .
`2x+3y=sinx`
Solution:
Here `2x+3y= sinx`
Differentiating both sides w.r. t.x, we have
`d/dx (2x+3y)`
`=d/dx(sinx)`
`2+3 dy/dx `
=`cosx`
`dy/dx=(cosx-2)/3`
Question: 2 .
`2x+3y=siny`
Solution:
Here `2x+3y=sin y`
Differentiating both sides w.r. t.x, we have
`d/dx(2x+3y)`
`d/dx(siny)`
`2+3 dy/dx`
`cosy dy/dx`
`(3-cosy)dy/dx`
`=-2=>dy/dx`
`=-2/(3-cosy)`
`=2/(cosy-3)`
Question: 3 .
`ax+by^2 =cosy`
Solution:
Here `ax+by^2=cosy`
Differentiating both sides w.r. t.x, we have
`d/dx(ax+by^2)`
`=d/dx(cos y)`
`a+2by dy/dx`
`=-siny dy/dx`
`(2by+siny) dy/dx`
` =-a=> dy/dx`
`=(-a)/(2by+siny)`
Question: 4 .
`xy+y^2=tanx+y`
Solution:
Here `xy+y^2=tanx+y`
Differentiating both sides w.r. t.x, we have
`d/dx(xy+y^2)`
`d/dx(tanx+y)`
`(x\ d/dx\y+d/dx\(x))+2y dy/dx`
`=secx^2+ dy/dx`
`x dy/dx +y+2y dy/dx`
`=sec^2 x+ dy/dx`
`(x+2y-1)dy/dx`
`=sec^2 x-y`
`dy/dx =(sec^2x-y)/(x+2y-1)`
Question: 5 .
` x^2+xy+y^2=100`
Solution:
Here ` x^2+xy+y^2=100`
Differentiating both sides w.r. t.x, we have
`d/dx (x^2+xy+y^2)`
`=d/dx (100)`
`2x+(x d/dx y+ y dy/dx x)+ 2y dy/dx=0`
`2x+x dy/dx+y+ 2y dy/dx=0`
`(x+2y)dy/dx`
`=-(2x+y)`
`dy/dx`
`= (-2x+y)/(x+2y)`
Question: 6 .
`x^3+x^2y+xy^2+y^3=81`
Solution:
Here `x^3+x^2y+xy^2+y^3=81`
Differentiating both sides w.r. t.x, we have
`d/dx(x^2+x^2y+xy^2+y^2)`
=`d/dx(81)`
`3x^2+(x^2 d/dx(y)+y d/dx (x^2))`
`+(x d/dx(y^2)+y^2 d/dx (x))`
`+3y^2 dy/dx=0`
`3x^2 + x^2 dy/dx+2xy+2xy dy/dx+y^2`
`+3y^2 dy/dx=0`
`(x^2+2xy+3y^2)dy/dx`
`=-(3x^2+2xy+y^2)`
`dy/dx=(-(3x^2+2xy+y^2))/(x^2+2xy+3y^2)`
Question: 7 .
`sin^2y + cos xy=pi`
Solution:
Here `sin^2y + cos xy=pi`
Differentiating both sides w.r. t.x, we have
`d/dx(sin^2y+cos xy)`
`=d/dx (pi)`
`2sin y d/dx (sin y)` `+(-sin xy). d/dx(xy)=0`
`2sin y cos y dy/dx -sinxy(x d/dx (y)+y d/dx(x))=0`
`2sin y cos y dy/dx-sin xy (x dy/dx +y)=0`
`sin 2y dy/dx -x sin xy dy/ dx -y sin xy =0`
`(sin 2y - x sinxy) dy/ dx =y sin xy`
`dy/dx = (ysin xy)/((sin 2y -x sin xy))`
Question: 8 .
`sin^2x+cos^2y=1`
Solution:
Here `sin^2x+cos^2y=1`
Differentiating both sides w.r. t.x, we have
`d/dx(sin^2+cos^2y)`
`=d/dx(1)`
`2sin x d/dx(sin x)` `+2 cos x d/dx (cos y)=0`
`2sin x cos x ` `+2 cos y.(-siny)dy/dx =0`
`sin 2x-sin2y dy/dx=0`
`=>dy/dx= (sin2x)/(sin2y)`
Question: 9 .
`y=sin^-1(2x/(1+x^2))`
Solution:
Here `y=sin^-1(2x/(1+x^2))`
Put `x=tan theta=> theta =tan^-1x`
`:. y= sin^-1((2tan theta)/(1+tan^2 theta))`
` =sin^-1(sin theta)=2theta`
`y=2tan ^-1 x`
Differentiating both sides w.r. t.x, we have
`dy/dx=d/dx(2tan^-1x)`
`=2xx 1/(1+x^2)= 2/(1+x^2)`
Question: 10 .
`y=tan^-1((3x-x^3)/(1-3x^2));-1/sqrt3< x <1/sqrt3`
Solution:
Here `y=tan^-1((3x-x^3)/(1-3x^2))`
Put `x=tan theta=> theta tan^-1x`
`y=tan^-1((3tan theta- tan^3theta)/(1-3tan^2theta))`
=`tan^-1(tan 3theta=3 theta`
`y=3tan^-1x`
Differentiating both sides w.r. t.x, we have
`dy/dx=d/dx (3tan^-1x)`
`=3xx 1/(1+x^2)=3/(1+x^2)`
Question: 11 .
`y=cos^-1((1-x^2)/(1+x^2));0< x <1`
Solution:
Here `y=cos^-1((1-x^2)/(1+x^2))`
Put `x=tan theta => theta =tan^-1x`
`:.y= cos^-1((1-tan^2 theta)/(1+tan^2 theta))`
`=1cos^-1(cos 2 theta)=2theta`
`y=2tan^-1x`
Differentiating both sides w.r. t.x, we have
`dy/dx= d/dx(2tan^-1x)`
`=2xx 1/(1+x^2)=2/(1+x^2)`
Question: 12 .
`y= sin^-1((1-x^2)/(1+x^2));0< x <1`
Solution:
Here `y=sin^-1((1-x^2)/(1+x^2));0< x <1`
Put `x=tan theta => theta =tan^-1x`
`:. y= sin^-1((1- tan^2 theta)/(1+tan^2 theta))`
=`sin^-1(cos2 theta)`
`sin ^-1[sin(pi/2-2theta)]`
`y= pi/2-2theta => y`
` = y=pi/2-2 tan^-1x`
Differentiating both sides w.r. t.x, we have
`dy/dx = d/dx[pi/2-2 tan^-1x]`
`=0-2xx 1/(1+x^2)`
`=- -2/(1+x^2)`
Question:13 . `y=cos^-1 (2x/(1+x^2));0 < x <1`
Solution:
Here `y=cos^-1 (2x/(1+x^2))`
put `x=tan theta => theta= tan^-1x`
`:. y=cos^-1((2tantheta)/(1+tan^2 theta))`
`= cos^-1 (sin 2theta)=cos^-1[cos(pi/2 -2theta)]`
`y= pi/2 -2theta `
`=>y= pi/2-2tan^-1 x`
Differentiating both sides w.r. t.x, we have
` dy/dx = d/dx[pi/2 - tan^-1x]`
`dy/dx=0-2xx1/(1+x^2)`
`=-2/(1+x^2)`
Question: 14 .
`y=sin^-1(2xsqrt(1-x^2)); -1 1/sqrt2 < x < 1/sqrt2`
Solution:
Here `y=sin^-1(2xsqrt (1-x^2))`
Put `x=sin theta => theta sin^-1 x`
`y=sin^-1[2sin theta sqrt(1-sin^2theta)]`
`=sin^-1 [2sintheta cos theta]`
`=sin^-1[sin2theta]=2theta`
`y=2 sin^-1 x`
Differentiating both sides w.r. t.x, we have
`dy/dx= d/dx(2sin^-1x)`
`=2xx (1/(sqrt(1-x^2)))`
`2/(sqrt(1-x^2))`
Question: 15 .
` y=sec^-1(1/(2x^2-1));0< x <1/sqrt2`
Solution:
Here ` y=sec^-1(1/(2x^2-1))`
Put `x=cos theta => theta= cos^-1x`
`:. y=sec^-1(1/(2cos^2 theta -1))`
`=sec^-1(1/(cos 2 theta))`
`y=2 cos^-1 x`
Differentiating both sides w.r. t.x, we have
`dy/dx= d/dx(2cos^-1x)`
=`2xx(1/(sqrt(1-x^2)))`
=`2/sqrt (1-x^2)`
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