Continuity & Differentiability NCERT Solutions
NCERT Exercise 5.2
Differentiate the functions w.r.t.x in Exercise 1 to 8
Question:1 . `sin (x^2+5)`.
Solution:
Let `y=sin(x^2+5)`
Differentiating both sides w.r.t.x, we have
`d/dx(y)=d/dx sin(x^2+5)`
`dy/dx=cos(x^2+5).d/dx(x^2+5)`
`=cos (x^2+5).2x`
`2xcos(x^2+5)`
Question:2 . `cos (sinx)`
Solution:
Let `y=cos(sinx)`
Differentiating both sides w.r.t.x, we have
`d/dx (y)=d/dx cos(sinx)`
`dy/dx=-sin(sinx).d/dx(sinx)`
`=-sin(sinx).cosx`
`=-cosx.sin(sinx)`
Question:3 . `sin(ax+b)`
Solution:
Let `y=sin(ax+b)`
Differentiating both sides w.r.t.x, we have
`d/dx(y)=d/dx sin(ax+b)`
`dy/dx=cos(ax+b.d/dx(ax+b)`
`=cos (ax+b).a`
`=a.cos(ax+b)`
Question:4 . `sec(tan(sqrtx))`
Solution:
Let`y=sec(tan(sqrtx))`
Differentiating both sides w.r.t.x, we have
`d/dx(y)= d/dx sec(tan(sqrtx))`
`dy/dx=sec(tan(sqrtx))` `.tan(tan(sqrtx)).d/dx tan(sqrtx))`
`=sec(tan(sqrtx)).tan(tan(sqrtx))` `.sec^2sqrtx.d/sx(sqrtx)`
`=sec(tan(sqrtx)).tan(tan(sqrtx))`.`sec^2sqrtx.1/2sqrtx`
`=(sec(tansqrtx).tan(tan(sqrtx)).sec^2sqrtx)/(2sqrtx)`
Question:5 . `(sin(ax+b))/(cos(cx+d)`
Solution:
Let `y=(sin(ax+b))/(cos(cx+d)`
Differntiating both sides w.r.t.x, we have
`d/dx(y)= d/dx[(sin(ax+b))/(cos(cx+d))]`
`dy/dx=(cos(cx+d).d/dx sin(ax+b)-sin(ax+b)d/dx cos (cx+d))/[cos(cx+d)]^2`
`=(cos(cx+d).cos(ax+b).a -sin(ax+b)-sin(cx+d).c)/[cos(cx+d)]^2`
`=(a cos(cx+d).cos (ax+b)+c sin(ax+b).sin(cx+d))/[cos (cx+d)]^2`
`=(acos(cx+d)cos(ax+b))/[cos(cx+d)]^2+(csin(ax+b)sin(cx+d))/[cos(cx+d)]^2`
`=a cos (ax+b)sec(cx+d)` `+ c sin(ax+b)tan(cx+d)sec(cx+d)`
Question: 6 . `cosx^2.sin^2(x^5)`
Solution:
Let `y=cos x^2-sin^2(x^5)`
Differentiating both sides w.r.t.x, we have
`d/dx(y) =d/dx[cos x^3. sin^2 (x^5)]`
`dy/dx=cos x^3.d/dx [sin^2(x^5)]` `+sin^2(x^5).d/dx(cos x^3)`
`cos x^3 .2 sin (x^5).d/dx sin(x^5)` `+sin^2(x^5).-sin x^3. d/dx (x^3)`
`=cos x^3 .2sin(x^5).cos(x^5)d/dx(x^5)-sin^2(x^5). sin x^3 .3x^2`
`=Cos x^3 .2 sin (x)^5. cos (x^5).5x^4-sin^2(x)^5. sin x^3. 3x^2`
`=10 x^4. sin(x)^5.cos(x^5)cos x^3-3 x^2.sin x^3.sin^2 (x^5)`
Question:7 . `2sqrt cot(x^2)`
Solution:
Let `y=2sqrt cot(x^2)`
Differentiating both sides w.r.t.x, we have
`d/dx(y) =d/dx[2sqrt cot(x^2)]`
`dy/dx=2. 1/(2sqrt(cot(x^2))) .d/dxcot (x^2)`
`=1/(sqrt(cot (x^2))` `.-cosec^2(x^2)d/dx(x^2)`
`=1/(sqrt cot(x^2)).-cosec^2(x^2).2x`
`=(-2xcosec^2(x^2))/(sqrt cot (x^2))`
Question:8 . `cos (sqrtx)`
Solution:
Let `y= cos(sqrtx)`
Differentiating both sides w.r.t.x, we have
`d/dx(y) = d/ dx [cos sqrtx]`
`-sinsqrtx. d/dx(sqrtx)`
`-sin sqrtx. 1/2sqrt x=(-sin sqrtx)/(2sqrtx)`
Question: 9 .Prove that the function `f` given by `f(x)=|x-1|,x in R` is not differentiable at `x=1`
Solution:
Here `f(x) =|x-1|`
`{(x-1, if x>=1), (1-x, if x<1):}`
L.H.D. at `x=1`
`lim_(h->0) (f(1-h)-f(1))/((1-h-1))`
`lim_(h->0) (1-(1-h)-(1-1))/(-h)`
`lim_(h->0)h/-h =-1`
R.H.D. at `x=1`
`lim_(h->0) (f(1+h)-f(1))/(1+h-1)`
`lim_(h->0) ((1+h)-1-(1-1))/h`
`lim_(h->0)h/h=1`
`:. `L.H.D. `!=` R.H. D.
Thus `f(x)` is not differentiable at `x=1`
Question: 10 .Prove that the greatest integer function defined by
`f(x)= [x]0<\x\<3` is not diffeentiableat `x=1` and `x=2`
Solution:
Here `f(x) = [x]`
R.H.d. at `x=1`
`lim_(h->0) (f(1+h)-f(1))/(1+h-1)`
`lim_(h->0) (1-1)/h`
L.H.D. at `x=1=1`
`lim_(h->0) (f(1-h)-f(1))/(1-h-1)`
`lim_(h->0) (0-1)/-h`
`lim_(h->0) 1/h`
= not defined.
`:.` L.H.D. `!=` R.H.D.
Thus `f(x)` is not differentiable at `x-1`
R.H.D. at `x=2`
`lim_(h->0) (f(2-h)-f(2))/(2+h-2)`
`lim_(h->0) (2-2)/h=0`
L.H.D. at `x=2`
`lim_(h->0) (f(2-h)-f(2))/(2-h-2)`
`lim_(h->0) (1-2)/-h`
`lim_(h->0) -1/-h`
`lim_(h->0) 1/h`
=not defined
`:.`L.H.D `!=` R.H.D.
Thus `f(x)` is not differentiable at `x=2`
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