Find the derivative of g(x) = sin^2 x + cos^2 x using the chain rule or the product rule.
2007-03-10 03:04:11 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
And please tell me HOW to get the answer too!!
2007-03-10 03:12:37 · update #1
since we have to apply the product rule ,therefore sin^2x=sinx.sinx and cos^2x =cos.x.cosx
so the differrentiation would be
dx=sinx .dsinx/dx+sinx .dsinx/dx+cosx.dcosx/dx+cosx.dcosx/dx
=sinx.cosx+sinx.cosx+cos(-sinx)+cosx(-sinx)
=2sinxcosx-2cosxsinx
=0
2007-03-10 03:41:08 · answer #1 · answered by suchi 2 · 1⤊ 3⤋
f (x) = x^3 - 12x + a million . . . the 1st spinoff set to 0 shows turning or table sure factors f ' (x) = 3x^2 - 12 3x^2 - 12 = 0 3 * (x + 2) * (x - 2) = 0 x = 2 ... x = - 2 . . . the 2d spinoff evaluated at x = 2 and -2 determines if those factors are min, max, or neither. f ' ' (x) = 6x f ' ' (2) = 6*2 = 12 <== constructive fee shows x=2 is an area minimum f ' ' (-2) = 6*(-2) = -12 <== destructive fee shows x=-2 is an area optimum a.) x = - 2 is a optimum, and x=2 is a minimum ... so x = - infinity to -2 is increasing x = -2 to +2 is lowering x = +2 to + infinity is increasing b.) f (-2) = (-2)^3 - 12*(-2) + a million = 17 f (2) = (2)^3 - 12*(2) + a million = - 15 c.) . . . the 2d spinoff set to 0 shows inflection factors, or the place concavity ameliorations 6x = 0 x = 0 <=== inflection factor x = - 2 is a optimum, so could be concave down concavity ameliorations on the inflection factor(s) ... so x = - infinity to 0 is concave down x = 0 to + infinity is concave up
2016-10-01 21:30:46 · answer #2 · answered by rocio 4 · 0⤊ 0⤋
answer for this problem in simplest way will be
g(x)=sin^2 x+cos^2 x=1
derivative of g(x)=0
by chain rule
derivative of g(x)= 2(sin x)*(cos x) + 2(cos x)(-sin x)
=0.
2007-03-10 05:46:23 · answer #3 · answered by Anonymous · 1⤊ 1⤋
Well Sin^2x can be seen as sin x*sinx. using the product rule we can say that u=sin and v=sinx so that
d(uv)/dx= u*dv/dx + v*du/dx
= sin x * cos x + sin x * cosx = 2 sinx cos x
Similarly for cos^2x
d(uv)/dx= cos x* -sinx + cos x* -sinx= - 2 sinx cos x
Deriative of g(x) = 2 sinx cos x - 2 sinx cos x = 0
An easier way to known this is to use the identity that sin^2 x + cos^2 x = 1. The deriative of 1 is zero.
2007-03-10 03:15:05 · answer #4 · answered by The exclamation mark 6 · 0⤊ 1⤋
If I understood your notation properly, the trigonometric identity sin^2x+cos^2x=1. the derivative is 0.
2007-03-10 05:15:13 · answer #5 · answered by Rob M 4 · 0⤊ 1⤋
g'(x) = 2 sin x cos x - 2 cos x sin x = 0
2007-03-10 03:08:13 · answer #6 · answered by floridagators519 2 · 1⤊ 2⤋
sin^2x is the same as (sin(x))^2, so you can re-write the entire problem as (sin(x))^2 + (cos(x))^2....you then bring the exponent down and use chain rule, so you get 2(sin(x))(cos(x)) + 2(cos(x))(-sin(x))...you bring out the negative on the sin(x) and it comes to be zero
2007-03-10 03:15:32 · answer #7 · answered by aclotm 2 · 0⤊ 2⤋
sin² x + cos² x = 1
so
g(x) = 1
the derivative of any number is 0
g'(x) = 0
2007-03-10 03:14:09 · answer #8 · answered by Anonymous · 1⤊ 2⤋
take the derivative of the sin of 2x and multiply it by the derivative of 2x. add it to the derivative of cos of 2x times the derivative of 2x.
g'(x)=2cos2x-2sin2x.
2007-03-10 03:12:43 · answer #9 · answered by taf_48fan 2 · 0⤊ 3⤋
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2007-03-10 03:06:54 · answer #10 · answered by Anonymous · 0⤊ 6⤋