Math Twelve

Continuity & Differentiability NCERT Solutions

NCERT Exercise 5.5 Q:10-18

Question: 10 .`x^(cos x)+(x^2 +1)/(x^2- 1)`

Solution:

Let `y=x^(cos x)+(x^2 +1)/(x^2- 1)`

put `u= x^(xcos x)` and `v=(x^2+1)/(x^2 -1)`

Then `y=u+v`

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

Now `u= x^(x cos x)`

Taking logarithm on both sides, we have

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

=`x cos xlogx`

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

`d/dx(log u)`

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

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

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

`(du)/dx = x^(x cos x)[cos x + log x(-xsin x +cos x)]`

`(du)/dx =x^(xcos\ x)[cos x+ (1+ log x)-xsin\ x log\ x]`

Also `v=(x^2+1)/(x^2-1)`

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

`(dv)/dx=d/ dx((x^2+1)/(x^2-1))`

`= ((x^2-1)d/dx (x^2+1)-(x^2+1)d/dx(x^2-1))/(x^2-1)^2`

`(2x(x^2-1)-2x(x^2+1))/(x^2-1)^2 `

`(2x^3-2x-2x^3-2x)/(x^2-1)^2`

` (-4x)/(x^2-1)^2`

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

`(dy)/dx = x^(x cos x)[cos x(1+log x)- xsin x log x)]- (4x)/((x^2-1)^2)`

Question: 11 . `(x cos x)^x + (x sin x)^(1/x)`

Solution:

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

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

Then ` y=u+v`

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

Now `u= (x cos x)^x`

Taking logarithim on both sides, we have

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

`= xlog(x cos x)`

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

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

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

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

`(du)/dx = (x cos x)^x[1/(cos x){x d/ dx cosx + cos x d/dx (x)}+log (x cos x)]`

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

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

Also `v= (x sin x)^(1/x)`

Taking logarithm on both sides, we have

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

`= 1/x log (x sin x)`

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

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

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

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

`(x sin x)^(1/x)[1/(x^2 sin x) {x d/dx (sin x) + sin x d/ dx (x)}-(log(x sin x))/(x^2)]`

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

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

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

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

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

Find `(dy)/dx` of the functions given in exercise 12 to 15

Question: 12 .Here `x^y + y^x =1`

Solution:

Let `y= x^y + y^x =1`

Put `u=x^y` and `v=y^x`

Then `u+v=1`

`=> (du)/dx + (dv)/dx=0`

Now `u=x^2`

Taking logarithm on both sides , we have

`log u=log x^y`

`log u= y log x`

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

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

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

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

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

ALso `v=y^x`

Taking logarithm on both sides, we have

`log v= log(y^x)`

`=xlog y`

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

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

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

`(dv)/dx= u[x. 1/y .(dy)/dx +log y.1]`

`=y^x[y/x dy/dx + log y]`

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

`x^y[y/x +log x (dy)/dx]` `+ y^x[x/y (dy)/dx+ log y]=0 `

`x^(y-1)+x^ylog x (dy)/dx` `+xy^(x-1)(dy)/dx + y^x log y =0`

` (x^y log x +xy^(x-1)) (dy)/dx `

`=-(x^(y-1)y+y^x log y)`

`(dy)/dx = ((x^(y-1)y+y^x log x))/((x^y logx+ xy^(x-1)))``

Question: 13 .`y^x=x^y`

Solution:

Here `y^x=x^y`

Taking logarithm on both sides , we have

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

`x log y= y log x`

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

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

`x.d/dx(log y)+ log y.d/dx(x)`

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

`x.1/y (dy)/dxlog y.1`

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

`(x/y -logx)(dy/dx)`

` = y/x -log y`

`((x-y log x)/y)(dy)/dx `

` =((y-x log y)/x)`

`:.(dy)/dx =(y(y-x log y))/(x(x-y log x))`

Question: 14 . `(cos x)^y =(cos y)^x`

Solution:

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

Taking logarithm on both sides , we have

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

`ylog cos x= x log cos y`

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

`d/ dx (ylog cos x)`

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

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

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

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

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

`y/(coos x).(-sin x)` `+ log cos x(dy)/dx`

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

`-y tan x+ log cos x (dy)/dx`

=`-x tan y (dy)/dx +log cos y`

`(log cos x + tan y)(dy)/dx`

`log cos y + y tan x`

`:. (dy)/dx = (log cos y+ y tan x)/(log cos x + x tan y)`

Question: 15 . `xy =e^(x-y)`

Solution:

Let `xy=e^(x-y)`

Taking logarithm on both sides , we have

`log (xy)=log e^(x-y)`

`logx+log y= (x-y) log e`

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

`d/ dx [logx+log y]`

`= d/dx (x -y)`

`1/x + 1/y . (dy)/dx = 1- (dy)/dx`

`(1/y +1)(dy)/dx = 1- 1/x`

`((1+y)/y)(dy)/dx = (x- 1/x)`

`(dy)/dx = ((x-1)/x)(y/(1+y))`

`(y (x-1))/(x (y+1))`

Question: 16 .Find the derivative of the function given by

`f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)` and hence find `f(1)`

Solution:

Here `f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)`

Taking logarithm on both sides, we have

`log f(x)=log [(1+x)(1+x^2)(1+x^4)(1+x^8)]`

`=log(1+x)+log (1+x^2)+log(1+x^4)+log(1+x^8)`

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

`d/dx [logf(x)]= d/dx [log (1+x)+log (1+x^2)+log(1+x^4)+log(1+x^8)]`

`1/f(x).f(x)=1/(1+x).d/dx(1+x)+ 1/(1+x^2).d/dx (1+x^2)+1/(x+x^4)d/dx(1+x^4)1/(1+x^8).d/dx (1+x^8)`

`f(x) = [1/(1+x) +(2x)/ (1+x^2) + (4x^3)/(1+x^4) + (sx^7)/ (1+x^8)]`

`f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)[1/(1+x) + 2x/(1+x^2)+4x^3/(1+x^4) + (8x^7)/(1+x^8)]`

`:. f(1)= (1+1)(1+1^2)(1+1^4)(1+1^8)[1/(1+1) + (2xx1)/(1+1^2) +(4xx1^3)/(1+1^4) +(8xx1^7)/(1=1^8)]`

`2xx2xx2xx2[1/2+2/2+4/2+8/2]`

`=16xx15/2 =8xx15=120`

Question: 17 .Differentiate `(x^2-5x+8)(x^3+7x+9)` in three ways mentioned below

(i)By using product rule

(ii)by expanding the product to obtain a single polynomial

(iii)by logarithmic differentiation.

Do they all give the same answer?

Solution:

Let `y= (x^2-5x+8)(x^3+7x+9)`

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

`(dy)/dx=d/dx[(x^2-5x+8)(x^3+7x+9)] `

`(x^2-5x+8)d/dx(x^3+7x+9)` `+(x^3+7x+9)d/dx(x^2-5x+8)`

`(x^2-5x+8)(3x^2+7)` `+(x^3+7x+9)(2x-5)`

`=3x^4+7x^2-15x^3-35x+24x^2` `+ 56+2x^4-5x^3-14x^2`

`-35x+18x-45`

`=5x^4-20x^3=45x^2-52x +11`

(ii)`y= (x^2-5x +8)(x^3+7x+9)`

`y= x^5+7x^3+9x^2-5x^4` `-35x^2-45x+8x^3-56x+72`

`y=x^5-5x^4+25^x3 -26x^2+ 11x + 72`

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

`(dy)/dx=d/dx(x^5-5x^4+15x^3-26x^2+11x+72)`

=`5x^4- 20x^3+ 45x^2-52x+11`

(iii)`y=(x^2-5x+8)(x^3+7x+9)`

Taking logarithm on both sides, we have

`log y =log[(x^2-5x+8)(x^3+7x+9)]`

`=log[(x^2-5x+8)+log(x^3+7x+9)]`

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

`d/dx (log y)=d/dx[log(x^2- 5x+8)+log (x^3+7x+9)]`

`1/y (dy)/dx = 1/(x^2-5x+8).d/dx (x^2-5x+8)` `+1/(x^3+7x+9).d/dx(x^3+7x +9)`

`(dy)/dx=y[(2x-5)/(x^2-5x+8)` ` +(3x^2 +7)/(x^3+7x+9)]`

`(dy)/dx=(x^2-5x+8)(x^3+7x+9)` `[((2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8))/((x^2-5x+8)(x^3+7x+9))]`

`2x^4+14x^2+18x-5x^3-35x-45+3x^4-15x^3_24x^2+7x^2-35x+56`

`5x^4-20x^3+45x^2-52x+11`

Question: 18 . If `u, v` and `w` are functions of `x` then show that

`d/dx (u.v.w)`

`(du)/dx v.w.u. du/dx. w+u.v (dw)/dx`

intwo ways - first by repeated applications of product rule, second by logarithmic differentiation.

Solution:

Let` y =u.v.w`

=`(u.v).w`

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

`(dy)/dx = d/dx[(u.v).w]`

=`u.v.d/dx(w)+w.d/dx(u.v)`

=`u.v.(dw)/dx +w[u.(dv)/dx+v.(du)/dx]`

=`u.v.(dw)/dx +u.(dv)/dx.w(du)/dx v.w`

=`(dw)/dx v.w+u.(dv)/dx.w+u.v.(dw)/dx`

Also `y=u.v.w`

Taking logarithm on both sides, we have

`log y=log(u.v.w)`

`=log u+logv+logw`

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

`d/dx (log y)`

`= d/dx[log u+log v+ log w]`

=`=1/y (dy)/dx=1/u .(du)/dx+1/v.(dv)/dx +1/w .(dw)/dx`

`(dy)/dx=y[1/u.(du)/dx +1/v.(dv)/dx +1/w. (dw)/dx]`

`=u.v.w[1/u.(du)/dx +1/v.(dv)/dx +1/w. (dw)/dx]`

`=(du)/dx u.w+u.(dv)/dx w.u.v. (dw)/dx]`

12-math-home


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