d/dx sin x = cos x

is there any proof for this?

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- May 6th 2010, 07:01 AMgrgrsanjayproof of derivative of sin(x)
d/dx sin x = cos x

is there any proof for this? - May 6th 2010, 07:03 AMgrgrsanjay
int sin x dx = -cos x + C

is there any proof for this too?

and why does always +c comes for any indefinite integration - May 6th 2010, 07:13 AMgrgrsanjay
i just got answer for one question

d/dx sin(x) = lim(d->0) ( sin(x+d) - sin(x) ) / d

= lim ( sin(x)cos(d) + cos(x)sin(d) - sin(x) ) / d

= lim ( sin(x)cos(d) - sin(x) )/d + lim cos(x)sin(d)/d

= sin(x) lim ( cos(d) - 1 )/d + cos(x) lim sin(d)/d

= sin(x) lim ( (cos(d)-1)(cos(d)+1) ) / ( d(cos(d)+1) ) + cos(x) lim sin(d)/d

= sin(x) lim ( cos^2(d)-1 ) / ( d(cos(d)+1 ) + cos(x) lim sin(d)/d

= sin(x) lim -sin^2(d) / ( d(cos(d) + 1) + cos(x) lim sin(d)/d

= sin(x) lim (-sin(d)) * lim sin(d)/d * lim 1/(cos(d)+1) + cos(x) lim sin(d)/d

= sin(x) * 0 * 1 * 1/2 + cos(x) * 1 = cos(x)

was my proof correct? - May 6th 2010, 07:47 AMHallsofIvy
That works fine once you have and .

Those often are deduced by a slightly "hand-waving" geometric argument, using the definition of sin(x) and cos(x) in terms of the unit circle.

It is also possible to**define**sin(x) to be and define cos(x) to be . It is not difficult to show that those series converge uniformly for all x and so we can differentiate "term by term". That is, .

Quote:

and why does always +c comes for any indefinite integration

(The other way can also be proved- that if F'(x)= f(x) and G'(x)= f(x), then F(x)- G(x) is a constant but that's a bit harder.) - May 6th 2010, 08:32 AMgrgrsanjaythanks
thank you very much