Math Twelve

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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