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प्रश्न
x cos x
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उत्तर
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`
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