Find out the derivtive of (sin x)/x using the first principle of derivative.

Dear Abhingya

let f(x) = (sinx)/x

so f(x+h) =(sin (x+h))/(x+h)

fo f'(x) = Lt _h→0 {f(x+h) -f(x)} /h

          = Lt _h→0 {(sin (x+h))/(x+h )  - (sinx)/x } /h

          =Lt _h→0 {xsin (x+h)   - (x+h)sinx } /hx(x+h)

          =Lt _h→0 {x(sin (x+h)   - sinx } /hx(x+h)    - Lt _h→0hsinx /hx(x+h)

          use sinC -sinD formula

          =Lt _h→0 {2(cos (x+h/2)  sin(h/2) } /h(x+h)    - Lt _h→0hsinx /hx(x+h)

         now open series for sin(h/2) and apply limit

 f'(x) = cos(x) /x  - sin(x) /x²

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Badiuddin

Dear Student,
Please find below the solution to your problem.

let f(x) = (sinx)/x
so f(x+h) =(sin (x+h))/(x+h)
fo f'(x) = Lt h→0 {f(x+h) -f(x)} /h
= Lt h→0 {(sin (x+h))/(x+h) -(sinx)/x } /h
=Lt h→0 {xsin (x+h) -(x+h)sinx } /hx(x+h)
=Lt h→0 {x(sin (x+h)- sinx } /hx(x+h) - Lt h→0hsinx /hx(x+h)
use sinC -sinD formula
=Lt h→0 {2(cos (x+h/2) sin(h/2) } /h(x+h) - Lt h→0hsinx /hx(x+h)
now open series for sin(h/2) and apply limit
f'(x) = cos(x) /x - sin(x) /x2

Thanks and Regards