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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 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards,Askiitians ExpertsBadiuddin
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
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation.
All the best. Regards,Askiitians ExpertsBadiuddin
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