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Question

Differentiate sin xx using the first principle.

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Solution

Let y=sin xx

Let δy be an increment in y, corresponding to an increment δx in x.

Then,y+δy=sin(x+δx)(x+δx)

δy={sin(x+δx)(x+δx)sin xx}={x sin (x+δx)(x+δx)sin x}x(x+δx)

δyδx={x sin (x+δx)(x+δx)sin x}x(x+δx).δx

dydx=limδx0 δyδx

=limδx0{x sin (x+δx)(x+δx)sin x}x(x+δx).δx

limδx0x[sin (x+δx)sin x](δx)sin xx(x+δx).δx =limδx0x[2cos(x+δx2)sin(δx2)(δx)sin x]x(x+δx).δx

={limδx0cos(x+δx2).limδx0sin(δx/2)(δ/2)limδx01(x+δx)limδx0sin xx(x+δx)}

=(cos x×1×1x)sin xx2=(cos xxsin xx2)=(x cos xsin x)x2

Hence,ddx(sin xx)=(x cos xsin x)x2


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