PROVE THAT PROBLEMS IN DIFFERENTIATION

Question 1 :

Differentiate the following

If u = tan^-1 [√(1+x²) - 1]/x and v = tan^-1x, find du/dv

Solution :

 v = tan^-1x

x  =  tan v

u = tan^-1 [√(1+tan²v) - 1]/tan v

u = tan^-1 [√(sec²v) - 1]/tan v

u = tan^-1 [sec v - 1]/tan v

By applying sec v = 1/cos v and tan v = sin v / cos v

u = tan^-1 ((1 - cos v)/sin v)

u = tan^-1 ((2sin² (v/2))/2sin (v/2) cos (v/2))

u = tan^-1 ((sin (v/2)/cos (v/2))

u = tan^-1 (tan (v/2))

u = v/2

du/dv  =  1/2

Question 2 :

Find the derivative with tan^-1 (sin x/(1 + cos x)) with respect to tan^-1 (cos x/(1 + sin x))

Solution :

Let u = tan^-1 (sin x/(1 + cos x))

u  =  tan^-1 [(2 sin (x/2) cos (x/2)/(2 cos² (x/2))]

  =  tan^-1 [(sin (x/2) /(cos (x/2)]

  =  tan^-1 (tan (x/2))

u = x/2

du/dx  =  1/2    ------(1)

Let v  =  tan^-1 (cos x/(1 + sin x))

v  =  tan^-1 (sin (π/2 - x)/(1 + cos (π/2 - x))

By applying the trigonometric formula for sin x and 1 + cos x, we get

v  =  tan^-1 (sin (π/2 - x/2)/cos (π/2 - x/2))

v  =  tan^-1 (tan (π/2 - x))

v  =  (π/2 - x/2)

dv/dx  =  -1/2   ------(2)

From (1) and (2), we have to find the value of du/dv

(1)/(2)  ==>  (1/2) / (-1/2)

  =  -1

Question 3. :

If y = sin^-1x then find y''

Solution :

y = sin^-1x

y'  =  1/√(1 - x²) 

√(1-x²)  (y')  =  1

√(1-x²)  (y'') + y' (1/2√(1-x²))(-2x)  =  0

√(1-x²)  (y'') + y' (-x/√(1-x²))  =  0

((1-x²)  (y'') - y' x)/√(1-x²)  =  0

(1-x²)  (y'') - y' x  =  0

y''  =  xy'/(1-x²)

y''  =  xy'/(1-x²)

By applying the value of y', we get

y''  =  x( 1/√(1 - x²))/(1-x²)

y''  =  x/(1 - x²)^3/2

Question 4 :

If y = e^(tan^-1x) show that (1+ x² ) y'' + (2x −1) y' = 0 .

Solution :

y = e^(tan^-1x) 

log y  =  tan^-1x

(1/y)y'  =  1/(1+x²)

y'  =  y/(1+x²)

y''  =  [(1 + x²)y' -  y(2x)] / (1+x²)²

(1+x²)² y''  =  [(1 + x²)y' -  y(2x)]

By dividing (1 + x²) on both sides, we get

  (1+x²) y''  =  y' - 2xy/(1+x²)

  (1+x²) y''  =  y' - 2xy'

  (1+x²) y''  =  y'(1 - 2x)

  (1+x²) y''  =  - y'(2x - 1)

  (1+x²) y'' + y'(2x - 1)   =  0

Hence proved.

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