Math Twelve

Continuity & Differentiability NCERT Solutions

NCERT Exercise 5.5

Differentiate the function given in Exercise 1 to 11 w.r.t. `x`.

Question: 1 .`cos x.cos 2x.cos3x`

Solution:

Let `y=cosx.cos2x.cos3x.`

Taking logarithm on both sides, we have.

`log y= log|cosx.cos2x.cos3x|`

`=log cos x` `+log cos 2x+logcos 3x`

Differentiating both sides w.r.t. `x`, we have

`d/dx(log y)=d/dx[log cos x+log cos 2x` `+logcos 3x]`

`1/y.dy/dx =[1/cos x. d/dx(cos x)` `+1/cos 2x.d/dx(cos 2x)+1/(cos 3x). d/dx (cos 3x)]`

`1/y.dy/dx =[1/cos x. (-sin x)` `+1/cos 2x.(-sin2x).d/dx(2x)+1/(cos 3x). (-sin 3x) .d/dx(3x)]`

`1/y.dy/dx` `=[-tanx-2 tan 2x -3 tan 3x]`

`dy/dx ` `=y[-tanx-2 tan 2x -3 tan 3x]`

`dy/dx` `=-(cos x cos 2x cos 3x)(tan x+2 tan 2x+ 3tan 3x)`

Question: 2 . `sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))`

Solution:

Let y=`sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))`

Taking longarithm on both sides, we have

`log y=log [sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))]`

=`1/2[log(x-1)+log(x-2)-log(x-3)-log(x-4)-log(x-5)]`

Differentiating both sides w.r.t. `x`, we have

`d/dx (log y)=d/dx {1/2[log(x-1)` `+log(x-2)-log(x-3)-log(x-4)-log(x-5)]}1/y. dy/dx`

`1/2[1/(x-1) d/dx(x-1)` `+1/(x-2) d/dx(x-2)(-1)/(x-3) d/dx (x-3) ` `(-1)/(x-4) d/dx(x-4)- 1/(x-5) d/dx (x-5)]`

`dy/dx=y/2[1/(x-1)+ 1/(x+2) - 1/(x-3) - 1/ (x-4) - 1/(x-5)]`

` dy/dx = 1/2sqrt(((x-1)(x-2))/((x-3)(x-4)(x-5)))[1/(x-1)+ 1/(x-2)-1/(x-3)- 1/(x-4)-1/(x-5)]`

Question: 3 .`(log x)^(cos x)`

Solution:

Let `y=(log x)^(cos x)`

Taking logarithm on both sides, we have

`log y= log[(log x)^(cos x)]`

`cos x log (log x)`

Differentiating both sides w.r.t. `x`, we have

`d/dx (log y) =d/dx [cos x log (log x)]`

`1/y. dy/dx = cos x. d/dx log (log x)` `+ log (log x).d/dx(cos x)`

`1/y. dy/dx = cos x. 1/log x d/dx (log x)` `+ log (log x).(-sin x)`

`dy/ dx =y[(cos x)/(log x) xx 1/x -sin x log (log x)]`

`dy/ dx =(log x)^(cos x)[(cos x)/(log x) -sin x log (log x)]`

Question: 4 . `x^x - 2^(sin x)`

Solution:

Let y =`x^x - 2^(sin x)`

Put `u=x^2` and `v= 2^(sin x)`

Then `y= u-v=> dy/dx = (du)/dx- (dv)/ dx`

Now `u=x^2`

Taking logarithim on both sides, we have

`log u= log(x^x)`

`log u = x log x`

Differentiating both sides w.r.t. `x`, we have

` d/dx (log u)= d/dx (x log x)`

` 1/u (du)/dx = x. d/dx (log x)` `+ log x d/dx (x)`

`(du)/ dx = u[x xx 1/x+ log x xx1]=x^x [1+log x]`

Also `v=2^(sin x)`

Taking logarithm on both sides, we have

`log v = log(2^(sin x))`

`=sin x log 2`

Differentiating both sides w.r.t. `x`, we have

`d/dx (log v)= d/dx(sin x log 2)`

`1/v dv/dx = cos x log 2`

`dv/dx = v [cos x log 2]`

` =2^(sin x) [cos x log 2]`

Putting value of `(du)/(dx)` and `(du)/(dx)` in (i), we get

`(dy)/(dx)=x^2[1+ log x]` `-2^(sin x)[cos x log 2]`

Question: 5 .

`(x+3)^2 .(x+4)^3 .(x+ 5)^4`

Solution:

Let y=

`(x+3)^2 .(x+4)^3 .(x+ 5)^4`

Taking logarithm on both sides, we have

`log y=log [(x+3)^2 .(x+4)^3 .(x+ 5)^4]`

`=log(x+3)^2 +log(x+4)^3 + log(x+ 5)^4`

`log y = 2 log (x+3)+3log(x+4)+4log(x+5)`

Differentiating both sides w.r.t. `x`, we have

`d/dx (log y)= d/dx [2log (x+3)` `+ 3 log (x+4) +4 log (x+5)]`

`1/y .dy/dx = [2. 1/(x+3) .d/dx(x+3)+ 3. 1/(x+4).d/(dx)(x+4) +` `4. 1/(x+5) d/dx (x+5)]`

` dy/dx= y[2/(x+3)+ 3/(x+4) 4/(x+5)]`

`(dy)/(dx) = (x+3)^2 .(x+4)^3(x+5)^4` `[2/(x+3)+ 3/(x+4) 4/(x+5)]`

Question: 6 . `(x+1/x)^x + x^((1+1/x))`

Solution:

Let y= `(x+1/x)^x + x^((1+1/x))`

Put `u=(x+1/x)^x and v= x^((1+1/x))`

Then `y=u+v`

`=>dy/dx = (du)/(dx)+ (dv)/(dx)`

Now `u=(x+1/x)^x`

taking logarithm on both sides, we have

`log u= log(x+1/x)^x`

`log u=xlog(x+1/x)`

Differentiating both sides w.r.t. `x`, we have

`d/dx(log u)`

`=d/dx[xlog(x+1/2)]`

`1/u.(du)/dx=x.d/dx log (x+1/x)` `+ log (x+1/x)d/dx (x)`

`(du)/dx=u[x. 1/(x+1/x) .d/dx (x+1/x) +log (x+1/x).1]`

`=u[x^2/(x^2+1).(1-1/x^2) + log (x+1/x)]`

`=(x+ 1/x)^2.[(x^2-1)/(x^2+1) + log (x^2+1/x)]`

Also `v=x^(1+1/x`

Taking logarithm on both sides, we have

`log v= x^(1+1/x)`

`log v= (1+1/x) log x`

Differentiating both sides w.r.t. `x`, we have

`d/dx (log v)=d/dx[(1+1/x)log x]`

`1/v dv/dx= (1+1/x).d/dx log x + log x(1+1/x)`

` dv/dx =v[(x+1)/x xx 1/x + log x xx (-1)/x^2]`

`x^(1+1/x) [(x+1)/x^2 - (log x)/x^2]`

=`x^(1+ 1/x)[(x+1-log x)/x^2]`

Putting value of `(du)/dx `and `(dv)/dx` in (i) we get

`(dy)/dx=(x+ 1/x)^2[(x^2-1)/(x^2+1) ` `+ log (x^2+1/x)]+x^(1+ 1/x)[(x+1-log x)/x^2]`

Question: 7 .`(log x)^2 + x^(log x)`

Solution:

Put `u=(log x^x` and `v=x^(log x)`

`y= u+v`

`=>(dy)/dx =(du)/dx + (dv)/dx`

Now `u = (log x)^x`

Taking logarithm on both sides, we have

`log u= log (log x)^x`

Differentiating both sides w.r.t. `x`, we have

`d/dx(log u)=d/dx[x log(log x)^x]`

`1/u(du)/dx = x. d/(dx) log (log x)+` ` log (log x).d/dx (x)`

`(du)/dx = u[x. 1/(log x) . d/ dx (log x)` ` + log (log x).1]`

`(du)/dx = (log x)^x[x. 1/(log x) . 1/x +log (log x)]`

`(du)/dx =(log x)^x[1/(log x) + log (log x)]`

`(log x)^(x-1) [1+log x. log (log x)]`

Also `v= x^(log x)`

Taking logarithm on both sides, we have

`log v= log(x^(log x))`

`log x. log x = (log x)^2`

Differentiating both sides w.r.t. `x`, we have

` d/dx (log v)= d/dx[(log x)^2]`

`1/v . (dv)/dx= 2log x. d/dx (log x)`

`dv/dx =v[2 log x .1/x]`

` x^(log x)[(2 log x)/x]`

`2x^(log x-1).log x`

putting value of `(du)/dx` and `(dv)/dx` in (i), we get

`dy/ dx =(log x)^(x-1[1=log x. log (log x)]` `+(2x^(log x-1).1. log x)`

Question: 8 .

`(sin x)^x + sin^(-1) sqrt x`

Solution:

Let` y=(sin x)^x + sin^(-1) sqrt x`

Put `u=(sin x)^x` and v= `sin^-1 sqrtx`

Then `y=u+v

`=> (dy)/dx = (du)/dx + (dv)/dx ----(i)`

Now `u= (sin x)^x`

Taking logarithm on both sides, we have

`log u=log (sin x)^x`

`=xlog sin x`

Differentiating both sides w.r.t. `x`, we have

`d/dx (log u)` `= d/dx[ x log sin x]`

`1/u (du)/dx ` `= x d/dx (log sin x)+log sin x d/dx(x)`

`(du)/dx =` `u[x. 1/(sin x) d/dx (sin x) + log sin x.]`

`(du)/dx =` `(sin x)^x[x. 1/(sin x) cos x+log sin x]`

`=(sin x)^x [x cot x+ log sin x]`

Also `V= sin^(-1)sqrt x`

Differentiating both sides w.r.t. `x`, we have

`(dv)/dx =d/dx (sin^-1 sqrtx)`

=`1/(sqrt(1-(sqrtx)^2)).d/dx(sqrtx)`

`= 1 /(sqrt1-x). 1/ (2sqrtx)`

=`1/(2sqrtx-x^2)`

Putting value of `(du)/dx` and `(dv)/dx` in (i), we get

`(dy)/dx = (sin x)^x [xcot x + log sin x]+ 1/(2sqrt(x-x^2)`

Question: 9 . `x^(sin x) + (sin x)^(cos x)`

Solution:

Let `y= x^(sin x) + (sin x)^(cos x)`

Put `u=x^(sin x)` and `v=(sin x)^(cos x)`

Then ` y=u+v `

`=> (dy)/dx=(du)/dx +(dv)/dx `

Now `u=x^(sin x)`

taking logarithm on both sides, we have

`log u= log x^(sin x)`

`=sin x log x`

Differentiating both sides w.r.t `x`, we have`

`d/dx (log u)`

`=d/dx[sin x log x]`

`1/u . (du)/dx`

` = sin x . d/ dx (log x)` `+ log x . d/dx (sin x)`

`(du)/dx = u[sin x . 1/x + . cos x]`

`(du)/dx = x^(sin x)[(sin x)/x` ` + cos x. log x]`

Also `v=(sin x)^(cos x)`

Taking logarithm on both sides, we have

log `u= log x^(sin x)`

`cos x log sin x`

Differentiating both sides w.r.t `x`, we have`

`d/dx (log v)`

=` d/dx (cos x log sin x)`

`1/v (dv)/dx`

`= cos x d/dx (log sin x)` `+ log sin x d/dx(cos x`

`(dv)/dx = v[cos x 1/(sin x).d/dx (sin x)` `+ log sin x.(-sin x)]`

`(dv)/dx =(sin x)^(cos x)[cos x cot x - sin x log sin x]`

putting value of `(du)/dx` and `(dv)/dx`in (i), we get

`(dy)/dx = x^(sin x)[(sin x)/x ` `+ cos x. log x]`

`+(sin x)^(cos x)[cos x cot x - sin x log sin x]`

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