Question

cos (x² + 1)

Answer 1

Let `f(x) = cos(x^2 + 1)`   .......(i)

⇒ `f(x + Δx) = cos[(x + Δx)^2 + 1]`  ......(ii)

Subtracting equation (i) from equation (ii) we get

`f(x + Δx) - f(x) = cos[(x + Δx)^2 + 1] - cos(x^2 + 1)`

Dividing both sides by Δx we get

`(f(x + Δx) - f(x))/(Δx) = (cos[(x + Δx)^2 + 1] - cos(x^2 + 1))/(Δx)`

⇒ `lim_(Δx -> 0) (f(x + Δx) - f(x))/(Δx) = lim_(Δx -> 0) (cos[(x + Δx)^2 + 1] - cos(x^2 + 1))/(Δx)`

f'(x) = `lim_(Δx -> 0) (cos[(x + Δx)^2 + 1] - cos(x^2 + 1))/(Δx)`  ......[By definitions of differentiations]

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

= `lim_(Δx - > 0) (sin[((x + Δx)^2 + 1 - x^2 - 1)/2])/(Δx)`    .....`[because cos C - cos D = - 2sin  (C + D)/2 * sin  (C - D)/2]`

`-2 sin [(x^2 + Δx^2 + 2x * Δx + x^2 + 2)/2]`

= `lim_(Δx -> 0) (-2 sin [x^2 + (Δx^2)/2 + x Δx + 1] sin[Δx (Δx + 2x)/2])/(Δx)`

`- 2sin[x^2 + (Δx^2)/2 + x Δx + 1]`

= `lim_(Δx -> 0) (sin [Δx (Δx + 2x)/2])/(Δx[(Δx + 2x)/2]) xx ((Δx + 2x)/2)`

= `lim_((Δx -> 0),(because  Δx [(Δx + 2x)/2] -> 0))  -2sin [x^2 (Δx^2)/2 + xΔx + 1] * (sin[Δx ((Δx + 2x))/2])/(Δx[(Δx + 2x)/2]) xx [(Δx + 2x)/2]` 

Taking limit, we have

= `-2 sin (x^2 + 1) * 1 * (x)`

= `- 2x sin(x^2 + 1)`  ......`[because  lim_(x -> 0) sinx/x = 1]`