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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