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प्रश्न
cos (x2 + 1)
उत्तर
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]`
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