Application of Derivatives NCERT Solutions
NCERT Exercise: 6.4
Question: (1)Using differentials, find the approximate value of each of the following upto 3 places of decimal.
(i)`sqrt 25.3`
Solution:
(i)Let `y= sqrtx`
`(dy)/(dx)=1/2sqrtx`
Let `x=25` and `x+Delta x = 25.3`
`:. Delta x= (x+ Delta x) - x`
` = 25.3- 25 = 0.3`
``Delta y= sqrt(x+Delta x)-sqrtx`
` =sqrt (253)-sqrt25`
`:. sqrt (253-5)`
` sqrt (253)`
`= 5+Delta y`
Since `dy` is approximately equal to `Delta y` and is given by
`dy = (dy)/(dx).Delta x `
` = 1/(2sqrt x) xx03`
`=(0.3)/(2sqrt (25))`
`=(0.3)/(2xx5) `
`=(0.3)/ (10) = 0.03`
`:. sqrt(253) = 5+0.03 =5.03`
Thus approximate value of ` sqrt (25.3)` is ` 55.03`
(ii) `sqrt (49.5)`
Solution:
Let `y=sqrt x`
`(dy)/(dx)= 1/(2sqrt x)`
Let `x= 49` and `x + Delta x = 49.5`
`:.Delta x =(x+Delta x) - x`
` = 49.5 -49 = 0.5`
`Delta y= sqrt(x+Delta x) - sqrt x`
`= sqrt(49.5)-sqrt (49) = sqrt (49.5-7)`
`:. sqrt (49.5)`
`=7+Delta y`
Since `dy` is approximately equal to `Delta y` and is given by
`dy= (dy)/(dx).Delta x`
`=1/(2sqrt x)xx 0.5`
`= (0.5)/2sqrt(49)`
`= (0.5)/(2xx7)`
`= 0.5/14 = 0.036`
` squrt(49.5)= 7+0.036`
` =7.036`
Thus approximate value of `sqrt(49.5)` is `7.036`
(iii)`sqrt (0.6)`
Solution:
Let `y=sqrt x`
`(dy)/(dx)= 1/(2sqrtx)`
Let `x= 0.64` and `x+ Delta x= 0.6`
` :. Delta x = (x+ Delta x)- x `
`=0.6-0.64=-0.04`
`Delta y = sqrt (x+Delta x)- sqrt x `
` sqrt (0.6) -sqrt(0.64)`
`= sqrt (0.6)- 0.8`
`:. sqrt (0.6)= 0.8 + Delta y`
Since `dy` is approximately equal to `Delta y` and is given by
`dy=(dy)/(dx).Delta x`
` 1/ (2sqrtx) xx (-0.04)`
`= (-0.04)/(2 sqrt(0.64))`
`(-0.04)/16 `
`-0.025`
`sqrt(0.6)= `
`:. 0.8 + 0.025 = 0.775`
Thus approximate value of `sqrt(0.6)` is ` 0.775`
(iv) `(0.009)^(1/3)`
Solution:
Let `y= x^(1/3)`
`(dy)/(dx) = 1/(3x^(2/3))`
Let `x=0.008` and `x + Delta x = 0.009`
`:. Delta x= (x+Delta x)-x`
` =0.009 - 0. 008 =0.001`
`Delta y= ( x+Delta x)^(1/3) -(x) ^(1/3)`
`=(0.009 )^(1/3) -(0.008)^(1/3)`
`= (0.009)^(1/3) -0.2`
`:. ( 0.009)^(1/3) = 0.2 + Delta y`
since `dy` is approximately equal to `Delta y` and is given by`
` dy = (dy)/(dx) Delta x`
`= 1/(3x)^(2/3) xx (0.001)`
` = (0.001)/(3(0.008)^(2/3)`
`= (0.001)/(0.12) =0.008`
` :. (0.009)^(1/3) `
`=0.2 + 0. 008 = 0.208`
`0.208`.(V) `(0.999)^(1/10)`
Solution:
Let `y=x^(1/10)`
`(dy)/(dx)= 1/(10x)^(9/10)`
Let `x= 1` and `x+ Delta x - 0.999`
`Delta x =(x+Delta x) - x`
` = 0.999 -1 =-0.001`
`Delta y= (x+Delta x)^(1/10) - (x)^(1/10)`
` -(0.999)^(1/10) - (1)^(1/10)`
`=(0.999)^(1/10) -1`
` :. (0.999)^(1/10) =1+ Delta y`
Scince `dy` approximately equal to `Delta y` and is given by
`dy=(dy)/(dx). Delta x.`
`=1/(10x)^(9/10) xx (- 0.001)`
` = -(0.001)/(10 xx(1)^(9/10) `
`( 0.999)^(1/10)= 1-0.0001=0.9999`
`:. ( 0.999)^(1/10)= 1-0.0001=0.9999`
Thus approximate value of `(0.999)^(1/10)` is `0.9999`.
(vi)`(15)^(1/4)`
Solution:
Let `y= x^(1/4)`
`(dy)/(dx) =1/(4x)^(3/4)`
Let `x = 16` and `x+ Delta x = 15`
` :. Delta x = (x+Delta x)-x`
` =15 -16 =-1`
` Delta y = (x+ Delta x)^(1/4) -(x)^(1/4)`
`= (15)^(1/4)-(16)^1/4 `
` =(15)^(1/4)-2`
`:. (15)^(1/4) = 2+ Delta y`
Since `dy` is approximately equal to `Delta y` and is given by
`dy =(dy)/(dx)Delta x `
` =1/(4x)^(3/4) xx (-1)`
` (-1)/(4xx(16)^(3/4)`
` =(-1)/(4xx8)`
` =-0.031`
`:. (15)^(1/4)= 2-0.031=1.969`
Thus approximate value of `(15)^(1/4)` is ` 1.969.
(vii) `(26)^(1/3)`
Solution:
Let `y= x^(1/3)`
`(dy)/(dx) = 1/(3x)^(2/3)`
Let `x = 27` and ` x+ Delta x =26`
` Delta x=(x+Delta x) -x`
` =26 -27 =- 1`
` Delta y= (x+ Delta x)^(1/3) -(x)^(1/3)`
` =(26)^(1/3)-(27)^(1/3)`
`=(26)^(1/3)-3`
` :.(26)^(1/3)= 3+Delta y`
Since `dy` is approximately equal to `Delta y` and is given by
`dy = (dy)/(dx) .Delta x `
`= 1/(3x)^(2/3) xx(-1)`
` (-1)/(3xx(27)^(2/3) `
` (-1)/(3xx9)`
`= -0.037`
`:. (26)^(1/3) = 3-0.037 =2963`
Thus approximate value of `(26)^(1/3)` is `2.963`.
(Viii) `(255)^(1/4)`
Solution:
Let `y = x^(1/4)`
`(dy)/(dx)=1/(4x)^(3/4)`
Let `x=256 ` and `x+Delta x =255`
`:. Delta x= (x+Delta x)-x`
` =255-256=-1`
`Delta y= (x+Delta x)^(1/4)-(x)^(1/4)`
` = (255)^(1/4) -(256)^(1/4)`
`=(255)^(1/4)-4`
`:.(255)^(1/4)=4+Delta y`
Since `dy` is approximately equal to `Delta y` and is given by
`dy = (dy)/(dx) Deltax = 1/(4x)^(3/4) xx (-1)`
`(-1)/(4 xx (256)^(3/4))`
`= (-1)/(4xx64)`
`=-0.004`
`:. (255)^(1/4) = 4-0.004 = 3.996`.
Thus approximate value of `(255)^(1/4)` is ` 3.996`.
(ix) `(82)^(1/4)`
Solution:
Let `y=x^(1/4)`
`(dy)/(dx) = 1/4^(3/4)`
Let `x= 81` and `x+ Delta x = 82`
`:. Delta x = (x+Delta x)-x`
` = 82 -81=1`
`Delta y=( x+ Delta x)^(1/4) -(x)^(1/4)`
`= (81)^(1/4)- (82)^(1/4)=(82)^(1/4)-3`
`:. (82)^(1/4)=3+Delta y`
Sinnce `dy` is approximately equal to `Delta y` and is gven by
`dy = (dy)/(dx). Delta x`
` = 1/ (4x)^(3/4) xx1`
` 1/(4xx(81)^(3/4))`
`1/(4xx27)= 0.009`
`:. (82)^(1/4)= 3+0.009= 3.009`
Thus approximate value of `(82)^(1/4)` is `3.009`
(x) `(401)^(1/2)`
Solution:
Let `y= x^(1/2)`
`(dy)/(dx) = 1/(2sqrt x)`
Let `x= 400` and `x+Delta x =401`
`:. Delta x= (x+Delta x)-x`
` =401 -400=1`
`Delta y= (x+Delta x)^(1/2)-(x)^(1/2)`
` (401)^(1/2)- (400)^(1/2)`
` =(401)^(1/2) -20`
` :. (401)^(1/2) = 20 + Delta y`
Since `dy` is approximately equal to `Delta y` and is given by
`dy = (dy)/(dx).Delta x`
`1/(2sqrt x).1= `
`= 1/(2sqrt 400)`
`=1/(2xx20)`
` = 0.025`
`:. (401)^(1/2)`
` 20 + 0.025 =20.025`
Thus approximate value of `(401)^(1/2)` is `20.025`
(xi) `(0.003)^(1/2)`
Solution:
Let `y= x^(1/2)`
`(dy)/(dx) = 1/(2sqrt x)`
Let `x=0.0036` and `x+Delta x= 0.0037`
`Delta x= (x+Delta x)-x `
`=0.0037 - 0.0036 = 0.0001`
` Delta y= (x+Delta x)^(1/2)-(x)^(1/2)`
` (0.00 37) ^(1/2)- (0.0036)^(1/2)`
` (0.0037)^(1/2)-0.06`
` (0.0037)^(1/2)`
`= 0 .06 + Delta y`
Since `dy` Is approximately equal to `Delta y` and is given by
`dy =(dy)/(dx).Delta x`
` = 1/(2 sqrt x).(0.0001)`
` (0.0001)/(2sqrt(0.00036))`
`(0.0001)/(2xx0.06) = 0..0008`
` (0.0037)^(1/2)`
`= 0.06+0.0008=0.0608`
Thus approximate value of `(0.0037)^(1/2)` is `0.0608`
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