Math Twelve

Continuity & Differentiability NCERT Solutions

NCERT Exercise 5.4

Differentiate the following w.r.t. `x`.

Question: 1 . `e^x/(sin x)`

Solution:

Let `=e^x/(sin x)`

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

`dy/dx= d/dx(e^x/(sin x))`

`=(sinxd/dx(e^x)-e^x d/dx (sin x))/(sin^2x)`

`=(sinx.e^x-e^x cosx)/(sin^2x)`

=`(e^x(sin x-cos x))/(sin^2x)`

Question: 2 . `e^(-1x)`

Solution:

let `y=e^(-1x)`

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

`dy/dx = d/dx e^(sin-1x)`

`=e^(sin-1x).d/dx (sin^-1x)`

`=e^(sin-1x)1/sqrt(1-x^2)`

Question:3 . `e^(x^3)`

Solution:

Let `y=e^(x^3)`

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

`dy/dx =d/dx e^(x^3)`

`=e^(x^3).d/dx(x^3)`

`=e^(x^3).3x^2`

`=3x^2.e^(x^3)`

Question: 4 . `sin(tan^-1 e^-x)`

Solution:

Let `y=sin (tan^-1 e^-x)`

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

`dy/dx = d/dx [sin(tan^-1 e^-x)]`

`= cos (tan^-1 e^-x)` `. d/dx(tan^-1 e^-x)`

`=cos (tan^-1 e^-x). 1/(1+(e^-x)^2). d/dx (e^-x)`

`=cos (tan^-1 e^-x)1/(1+e^-2x).d/dx(-x)`

`=cos (tan^-1 e^-x). 1/(1+e^-2x) . e^-x.(-1)`

`=(-e^-x cos (tan^-1 e^-x))/(1+e^-2x)`

Question: 5 . `log (cos e^x)`

Solution:

Let `y=log(cose^x)`

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

`dy/dx=d/dx[log(cos e^x)]`

`=1/(cos e^x). d/dx (cos e^x)`

`=1/(cos e^x). -sin e^x.d/dx (e^x)`

`=1/(cos e^x). -sin e^x. e^x`

`=(-e^x sin e^x)/(cos e^x)`

`=-e^x tan e^x`

Question:6 .

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

Solution:

Let y= `e^x+e^(x^2)+e^(x^3)+...+e^(x^5)`

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

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

`=e^x+e^(x^2).d/dx(x^2)+e^(x^3) .d/dx (x^3)+e^(x^4).d/dx(x^4)+e^(x^5).d/dx (x^5)`

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

`= e^x+2xe^(x^2)+3x^2 e^(x^3)+ 4 x^3 e^(x^4)+5x^4e^(x^5)`

Question: 7 .`sqrt(e^(sqrtx)), x > 0`

Solution:

Let `y=sqrt(e^(sqrtx)), x > 0`

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

`dy/dx= d/dx (sqrt(e^(sqrtx)))`

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

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

=`1/(2sqrt(e^(sqrtx))) . (e^sqrtx).1/(2sqrtx)`

=`(e^(sqrtx))/(4sqrtx sqrt(e^(sqrtx)))`

`=(e^(sqrtx))/(4sqrt(x e^sqrtx))`

Question: 8 .`log (log x), x > 1`

Solution:

Let `y= log (log x)`

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

`dy/dx= d/dx[log(log x)]`

`=1/(log x .d/dx (log x)`

`=1/ log x. 1/x `

`=1/(x log x)`

Question: 9 . `(cosx)/(log x),x > 0`

Solution:

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

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

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

`=(log x d/dx (cos x)-cosx d/dx (log x))/(logx^2)`

`=(logx (-sinx)-cosx 1/x)/((logx)^2)`

`=(-sinx log x- (cos x)/x)/(log x)^2`

`=(-(xsinxlogx+cosx))/(xlogx)^2`

Question: 10 .`cos (log x+ e^x), x > 0`

Let `y= cos (logx+e^x)`

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

`dy/dx = d/ dx [cos(lod x+ e^x)]`

`=-sin(logx+e^x).d/dx (log x+e^x)`

`=-sin(logx+e^x).(1/x +e^x)`

`=-(1/x + e^x)sin(log x+ e^x)`

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