Continuity & Differentiability NCERT Solutions
Miscellaneous Exercise
Differentiate the functionsw.r.t. `x` in Exercise 1 to 11
Question: (1) `(3x^2-9x + 5)^9`
Solution:
Let ` y= (3x^2-9x+ 5)^9`
Differentiating boh sides w.r.t `x` we have
`(dy)/(dx) = d/(dx) (3x^2-9x+5)^9`
`= 9(3x^2-9x + 5)^8 d/(dx)(3x^2-9x+5)`
`9(3x^2- 9x +5) (6x-9)`
` =27(2x -3) (3x^2 - 9x =5)^3`
Question: (2) `sin ^2 x+ cos ^6x`
Solution:
Let ` y= sin^2x + cos^6 x`
Differentiating boh sides w.r.t `x` we have
`(dy)/(dx) = d/(dx) (sin^3 x + cos^6 x)`
`=3 sin^2 x . d/dx (sin x)`
` + 6 cos^5 x\ d/dx (cos x)`
`=3 sin^2x cos x + 6 cos^5 x(-5sin x)`
Question: (3) `(5x)^(3cos 2x)`
Solution:
Let` y = (5x)^(3cos 2x)`
Taking logarithim on both sides, we have
`log y = log (5x)^(3cos 2x)`
`= 3cos 2x log (5x)`
Differentia ting both sides w.r.t. `x` , we have
`d/dx (log y)` `= d/dx [3 cos 2x log (5x)]`
`1/y dy/ dx`
` =3 cos 2x . d/ dx [log (5x)]` `+log(5x) .d/ dx (3 cos 2x)]`
`(dy)/dx = y [3 cos 2x . 1/(5x) . d/dx (5x)+ log (5x) .- 3sin 2x . d/dx (2x)]`
`=(5x)^(3cos 2x)[3 cos 2x. 1/(5x) .5- log (5x). 3 sin 2x .2]`
`(5x)^(3 cos 2x)[3 cos 2x. 1/(5x) . 5- log(5x). 3sin 2x .2]`
Question: (4) ` sin ^-1 (xsqrt x), 0<= r<= 1`
Solution:
Let `y= sin^-1 (x sqrt x)=sin^-1 (x^(3//2))`
Differentiating both sides w.r.t. `x` we have
`dy/dx = d/dx[ sin^-1(x(3//2)]`
`=1/(sqrt(1-(x(3//2))^2)\ .d/dx (x^(3//2))`
`1/(sqrt(1-x^3))\ . 3/2 x^(1//2)`
`3/2 (sqrt x)/(sqrt(1-x^3)`
`3/2 sqrt(x/(1-x^3))`
Question: (5) `(cos ^-1 \ x/2)/(sqrt (2x+7)), -2< x < 2`
Solution:
Let `y= (cos ^-1 \ x/2)/(sqrt (2x+7))`
Differentiating both sides w.r.t. `x` , we have
`dy/dx = d/dx [(cos ^-1 \ x/2)/(sqrt (2x+7))]`
`= (sqrt(2x+7)d/dx (cos^-1\ x/2)-cos^-1(x/2).d/dx (sqrt (2x+7))) /((sqrt(2x+7)^2))`
`=(sqrt (2x+7). -1/(sqrt (1-(x/2)^2)) \. d/dx (x/2)- cos^-1 (x/2). 1/(2sqrt(2x+7))d/dx (2x+7))/(2x+7)`
`((sqrt (2x+7))/(2sqrt(1-x^2/4))- (cos^-1\ x/2)/(sqrt(2x+7)))/((2x+7))`
`=(-sqrt(2x+7))/(sqrt(4-x^2)(2x+7)) \- (cos^-1\ x/2)/(sqrt(2x+7)(2x+7))`
`= - [1/(sqrt(4-x^2)sqrt (2x+7))\+ (cos^-1 \ x/2)/((2x+7)^(3//2))]`
Question: (6) ` cot^-1 [(sqrt(1+sinx)+ sqrt(1-sin x))/(sqrt(1+sinx)- sqrt(1-sin x))], 0< x < pi/2`
Solution:
Let `y= cot^-1 [(sqrt(1+sinx)+ sqrt(1-sin x))/(sqrt(1+sinx)- sqrt(1-sin x))]`
` sqrt(1+sin x)= sqrt(sin^2 \ x/2 + cos^2 \ x/2 + 2 sin \ x/2 cos \ x/2)`
`sqrt ((sin \ x/2 + cos \ x/2)^2)`
`= sin \ x/2 + cos \ x/2`
` sqrt(1-sin x)= sqrt(sin^2 \ x/2 + cos^2 \ x/2 - 2 sin \ x/2 cos \ x/2)`
`sqrt ((sin \ x/2 - cos \ x/2)^2)`
`= sin \ x/2 - cos \ x/2`
` :. y= cot^-1 [(sin \ x/2 + cos \ x/2 + sin \ x/2 -cos \ x/2)/ (sin \ x/2 + cos \ x/2 - (sin \ x/2 -cos \ x/2))]`
`cot^-1 [(2sin \ x/ 2)/(2cos \ x/ 2)]`
`= cot^-1[tan \ x/2]= cot^-1[cot( pi/2 - x/2)] = pi/ 2 - x/2`
`:. y= pi/ 2 - x/ 2`
Differentiating both sides w.r.t. `x` , we have
` dy/dx = d/ dx ( pi/2 - x/2)= 1/2`
Question: (7) `(log x)^(log x), x > 1`
Solution:
Let `y= (log x)^(log x)`,
Taking logarithm on both sides, we have
`log y= log [(log x)^(log x)]`
`= log x log (log x)`
Differentiating both sides w.r.t. `x` , we have
` d/dx (log y)= d/ dx[log x. log(log x)]`
`1/y dy/dx = log x . d/ dx log (log x)+ log (log x).d/dx (log x)`
`= dy/dx =y[ log x. \ 1/log x . d/dx (log x)+log (log x)1/x ]`
`dy/dx =y [1/x +1/x log (log x)]`
`dy/dx =(log x)^(log x)[1/x + (log(logx)/x)]`
Question: (8) ` cos (a cos x+b sin x)` , for some constants a and b
Solution:
Let ` y= cos (a cos x+ bsin x)`
Differentiating both sides w.r.t. `x` , we have
` dy/dx= d/dx[cos (a cos x + b sin x)]`
`=- sin (a cos x+ b sin x) . d/dx (a cos x+ b sin x)`
`=- sin (a cos x+ b sin x).(-a sin x+ b cos x)`
` (a sin x -b cos x) . sin (a cos x+ b sin x)`
Question: (9) `(sin x - cos x )^(sinx-cos x) , pi/4 < x < (3pi)/4`
Solution:
Let `y= (sin x - cos x )^(sinx-cos x)`
Taking loagarithm on both sides, we have
`log y= log (sin x - cos x )^(sinx-cos x)`
`=(sin x - cos x )log(sinx-cos x)`
Differentiating both sides w.r.t. `x` , we have
` d/dx(log y)= d/dx[sin x-cos x).log (sin x -cos x)]`
`1/y dy/dx = (sin x- cos x). d/ dx log (sin x- cos x) + log (sin x-cos x). d/dx (sin x- cos x)`
`dy/dx =y [ (sin x -cos x) . 1/((sin x- cos x)) d/ dx ( sin x - cos x) + log (sin x- cos x) .(cos x- sin x)]`
`dy/dx =y [cos x+ sin x + (cos x+ sin x) log (sin x- cos x)]`
` dy/dx= (sin x- cos x)^(sin x- cos x) (cos x + sin x) [1= log ( sin x - cos x)]`
Question: (10) ` x^x + x^a + a^x + a^a`, for some fixed `a > 0` and `x > 0`
Solution:
Let ` y=x^x + x^a + a^x + a^a`
`y = e^(x log x)+ x^a + a^a + a^a` `[:. x^x= e^(xlog x)]`
Differentiating both sides w.r.t. `x` , we have
` dy/dx= d/dx[e^(x log x)+ x^a+a^x+ a^a]`
`=e^(x log x). d/dx(x log x)` `+ax^(a-1)+a^ x log a + 0`
`e^(x log x)[x d/dx (log x) + log x . d/dx (x)]` `+ ax^(a-1)+ a^x log a`
`= e^(x log x) [x. 1/x + log x.1] ` `+ ax^(a-1)+a^x log a`
`= e^(x log x)[ 1+ log x]` ` + ax^(a-1) + a^x log a`
` = x^x [ 1+ log x] ` `+ ax^(a-1)+ a^x log a`
Question: (11) ` x^(x^x-3)+ (x-3)^(x^2)`, for `x > 3`
Solution:
Let `y= x^(x^x-3)+ (x-3)^(x^2)`
Put `u= x^(x^2-3) and v= ( x- 3)^ (x^2)`
Then `y= u= v`
` => (dy)/(dx)=(du)/(dx) + (dv)/(dx)`
Now `u = x^(x^2-3)`
Taking logarithm on both sides, we have
`log u= log x^(x^2-3)`
`= (x^2-3) log x`
Differentiating both sides w.r.t. `x` , we have
` d/dx(log u)= d/dx[(x^2-3). log x]`
`1/u (du)/dx`
` =(x^2 -3) d/ dx (log x)` `+ log x . d/dx (x^2 -3)`
` =(du)/dx = u[(x^2-3)/x 2x log x]`
` (du)/dx =x^(x^2-3)[(x^2-3)/x + 2x log x]`
` Also v= (x-3)^(x^2)`
Taking logarithm on both sides, we have
`log v = (x-3)^(x^2)`
` = x^2 log (x-3)`
Differentiating both sides w.r.t. `x` , we have
` 1/v (dv)/dx= d/dx[x^2 log (x-3)]`
`(dv)/dx`
`=v[x^2 .d/ dx log (x-3)` `+log (x-3). d/dx (x^2)]`
` = (x-3)^(x^2)[ (x^2)/(x-3)` `+2x log (x-3)]`
putting value of ` (du)/dx` and `(dv)/dx` in (i) , we get
`(dy)/dx = x^(x^2-3)[(x^2-3)/x + 2x log x]`
` +(x-3)^(x^2)[ (x^2)/(x-3)] + 2x log (x-3)]`
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