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use the chain rule and product rule: f(x)=1/u(x), f'(x)=-1/u2(x)*u'(x) and f(x)=a(x)b(x), f'(x)=a'(x)b(x)+b'(x)a(x)
the derivative of sin is cos, and the derivative of cos in negative sin.
so f(x)=1/(sin(x)cos(x)), f'(x)=-1/(sin(x)cos(x))2*(cos2(x)-sin2(x))
or assume sin(x)cos(x)=.5sin(2x)
so using chain rule, f'(x)=-2/sin2(2x)*.5cos(2x)*2. Both answers are equivalent.
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What is the derivative of 1 divided by sinx?
y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx
What is sin 3x in terms of sin x?
given the identity sin(x+y)=sinx cosy + siny cosxsin2x = 2 sinx cosx andsin(2(x)+x) = sin 2x cos x + sinx cos 2xusing the last two identities givessin3x= 2 sinx cosx cosx + sinx cos2xfactoring the sinx we havesin3x = sinx(2cosx cosx+cos2x)which satisfies the requirement.However, we can simplify further since cos 2x = cosx cosx - sinx sinx (a well known identity)sin3x = sinx (2cosx cosx +cosx cosx - sinx sinx)so sin3x= sinx(3cosx cosx - sinx sinx)or sin 3x = 3.cos²x.sinx - sin³x* * * * *Good, but not good enough. The answer was required in terms of sin, not a mixture of sinx and cosx. Easily recitified, though, since cos²x = 1 - sin²xTherefore sin3x = 3*(1-sin²x)*sinx - sin³x= 3sinx - 3sin³x - sin³x= 3sinx - 4sin³x
When does cosx times sinx times sinx equal 1?
at the angles 0 and 360 degrees, or 0 and 2pi
What is the differentiation of sin x over 1 plus cosx?
The derivative is 1/(1 + cosx)
What is the value of sinX wrt cosX?
sin(x) = sqrt[ 1 - cos2(x) ]
