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Differential Calculus

Scorching goIITian

 Joined: 25 Dec 2010 Post: 241
25 Dec 2010 12:10:27 IST
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differentiation of sinx
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Differential Calculus

differentiation of sinx

Blazing goIITian

Joined: 8 Oct 2008
Posts: 8064
25 Dec 2010 12:11:10 IST
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it is cosx.........derived from a formula

New kid on the Block

Joined: 10 Jan 2011
Posts: 28
13 Jan 2011 18:18:48 IST
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Though u learn it as a formula .If u differentiate the sin x Taylor series expansion u get back the cos x Taylor series expansion.

$\sin\left( x \right) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!}.\!$

$\cos x = 1 - {x^2 \over 2!} + {x^4 \over 4!} - \cdots\!$

Blazing goIITian

Joined: 2 Sep 2007
Posts: 383
13 Jan 2011 19:47:47 IST
1 people liked this

you can solve it using first principles of differnciation

.

Blazing goIITian

Joined: 8 Sep 2008
Posts: 345
13 Jan 2011 20:18:23 IST
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Hope you know the first principle of differentiation..........

Anyways we'll take it one more time.....it is

f'(x) = Lt (h->0 )   [  f(x+h)-f(x) ] /  h

You can drive this equation from the slope formula ie slope m = y2-y1 /  x2-x1              in the above eq. h is x2-x1 and y2 - y1 is f(x+h) - f(h) and f'(x) is the slope of curve  ie dy/dx or 'm'........

Coming to ur ques.  now substitute sin  for f in the abov eq.....

Lt h->0  sin(x+h)-sinx  /  h

but sin(x+h)-sinx =  2 cos(x + h/2) sin (h/2)

now substituing the limits   ie sin h/2 /h/2  =1

and

then h=0 gives "cosx"  as the answer ie the derivative

New kid on the Block

Joined: 17 Dec 2010
Posts: 23
13 Jan 2011 21:06:23 IST
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friend, it is cos X . Other wise if u have a doubt then solve it using 1st principle. Its simple to solve.

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