cosx.cos2x.cos4x.cos8x.cos16x
cosx.cos2x.cos4x.cos8x.cos16x
Harshit Sahu , 15 Years ago
Grade 11
Differential Calculus> find the differential...
3 Answersvikas askiitian expert
Y = cosxcos2xcos4xcos8xcos16x
taking log both sides
logY = log(cosx.cos2x.cos4x.cos8x.cos16x)
logab = loga + logb (property)
logY =[ logcosx + logcos2x + logcos4x + logcos8x + logcos16x ] ...................1
differentiating wrt x
differentiatn of log(f(x) = f1x/ f(x)
(1/Y) . (y1) =
[ -sinx/cosx + -sin2x/2cos2x - sin4x/4cos4x - sin8x/8cos8x -16sin16x/16cos16x]
y1 = -Y[tanx + tan2x/2 + tan4x/4 + tan8x/8 + tan16x/16] .....................2
now put the value of Y from 1 in eq 2 u will get the desired result...
Akarsh Gaurav
y=f(x)=cosx.cos2x.cos4x.cos8x.cos16x
=2sinx.cosx.cos2x.cos4x.cos8x.cos16x/2sinx [
multiply 2sinx/2sinx in RHS]
=2sin2x.cos2x.cos4x.cos8x.cos16x/2.2sinx [{sin2x=2sinx.cosx} & multiply 2/2 in RHS]
SIMILARLY,
=2sin4x.cos4x.cos8x.cos16x/2.4sinx
=2sin8x.cos8x.cos16x/2.8sinx
=2sin16x.cos16x/2.16sinx
=sin32x/32sinx
differentiate both sides w.r.to x,
dy/dx=d(sin32x/32sinx)/dx
differentiate it you will getthe answer.
THANK YOU
Nitish jha
The question can be written asCos2^0x.cos2^1x.cos2^2x.cos2^3x.cos2^4xHere n=4Simply put in formulaSin2^(n+1)x/2^(n+1)sinxSin2^{5}x/2^5sinxSin32x/32sinxIf u want to find exact vakue of x, differentiate y w.r.t x
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