Please solve it and also explain.
2007-10-25 00:36:40 · 14 answers · asked by Maham 1 in Science & Mathematics ➔ Mathematics
d/dx(sinx.cosx)
=sinx.d/dx(cosx) + cosx.d/dx(sinx)
=sinx(-sinx) + cosx(cosx)
=cos2x
2007-10-25 00:50:12 · answer #1 · answered by Adi 1 · 1⤊ 0⤋
D Dx Sinx Cosx
2016-12-15 03:15:53 · answer #2 · answered by ? 4 · 0⤊ 0⤋
The given problem may be solved in the following two ways.
First Method - We have
Sinx. Cosx = 1/2 [ 2 Sinx.Cosx ] = 1/2 [ Sin 2x ]
Hence d/dx [ Sinx.Cosx ] can be written as d/dx [ 1/2 Sin2x ]
= 1/2 d/dx (Sin2x)
= 1/2 [ Cos2x . d/dx (2x) ]
= 1/2 [ (Cos2x) . 2 ] Since d/dx (2x) = 2
= (1/2).(2).Cos2x
= Cos2x ............... Answer
Second Method We have
d/dx (Sinx. Cosx) = Sinx. (-Sinx) + Cosx. Cosx
= Cos^2x - Sin^2x
= Cos2x ............. Ans
2007-10-25 02:18:15 · answer #3 · answered by Pramod Kumar 7 · 1⤊ 0⤋
Use the product rule: Derivative of the first function times the second function plus the derivative of the second function times the first function:
cosxcosx -sinxsinx=1-2sinxsinx since
cosxcosx + sinxsinx = 1
2007-10-25 00:51:15 · answer #4 · answered by oldschool 7 · 0⤊ 0⤋
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Do you mean derivative? You need to apply the chain rule. Let u = 3sinx - 8; then, y = ln u. dy/dx = (1/u) du/dx. From the expression for u above, du/dx = 3cosx. So, dy/dx = (1/3sinx - 8)*( 3cox) = (3cosx)/(3sinx - 8)
2016-04-04 21:51:44 · answer #5 · answered by Anonymous · 0⤊ 0⤋
d(sinxcosx)/dx
=sinx(-sinx)+cosx(cosx)
=cos^2 x - sin^2 x
=1- 2sin^2 x
use of product rule
2007-10-25 00:45:42 · answer #6 · answered by pigley 4 · 0⤊ 1⤋
sinx.cosx
= 2 sinxcosx /2
= 1/2*sin2x
its derivative will be
1/2*cos2x*2
= cos2x
2007-10-25 20:18:11 · answer #7 · answered by gauravragtah 4 · 0⤊ 1⤋
use the product rule.
sinxcosx d/dx = cosx(cosx) + (-sinx)(sinx) = cos^2x - sin^2x
= cos2x
2007-10-25 00:45:02 · answer #8 · answered by Linda K 5 · 1⤊ 0⤋
sinx cosx = (1/2)* sin2x
=> ( sinx cosx )' = (1/2)*2*cos(2x)
= cos 2x
2007-10-25 00:48:56 · answer #9 · answered by namvt2000 6 · 0⤊ 0⤋
d(sinxcosx) = cos(cosx) + sinx(-sinx)
= cos^2x -sin^2x
= cos^2x - (1- cos^2x)
= 2cos^2x -1
= cos2x
2007-10-25 00:48:23 · answer #10 · answered by Synchronizers 3 · 1⤊ 0⤋
sinx*cosx = cosx/2
or
cosx * 2
example
sin30 * cos30 = cos30/2
sin45*cos45 = cos45 * 2
sin60 * cos60 = cos60/2
sin35*cos35= cos35/2
thanx
Actually this is approximate answer for further solution
Reality is that , Sinx*Cosx is itself is simplified form.
Derivation is always we use to simplify the problem not to make it more complex.
We can make it as much complex as we can. But it's simple
2007-10-25 00:56:35 · answer #11 · answered by piyush 2 · 0⤊ 3⤋