Question

x cos x

Answer 1

Let `y = x cos x`  ......(i)

`y + Δy = (x + Δx) cos(x + Δx)`  ......(ii)

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

`y + Δy - y = (x + Δx) cos(x + Δx) - x cos x`

⇒ `Δy = x cos (x + Δx) + Δx cos (x + Δx) - x cos x`

Dividing both sides by Δx and take the limits,

`lim_(Δx -> 0) (Δy)/(Δx) = lim_(Δx -> 0) (x cos (x + Δx) - x cos x + Δx cos (x + Δx))/(Δx)`

`(dy)/(dx) = lim_(Δx -> 0) (x[cos(x + Δx) - cos x])/(Δx) + lim_(Δx -> 0) (Δx cos(x + Δx))/(Δx)`   ......`["By defination"  lim_(Δx -> 0) (Δy)/(Δx) = (dy)/(dx)]`

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

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

∴ `(Δx)/2 -> 0` Taking the limits, we have

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

= `- x sin x + cos x`