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Differential Calculus

Scorching goIITian

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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

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Sagar Saxena

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25 Dec 2010 12:11:10 IST

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it is cosx.........derived from a formula

Siddharth Mandal Mandal

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13 Jan 2011 18:18:48 IST

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cos (x) is the answer.

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.

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

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

NANDINI

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13 Jan 2011 19:47:47 IST

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you can solve it using first principles of differnciation

answer is cos x

carpe dium

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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'........

for More info read ur txt books or ping people here again............

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

pankajmahato

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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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