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Harshit Sahu Grade: 11
`        cosx.cos2x.cos4x.cos8x.cos16x`
6 years ago

## Answers : (3)

vikas askiitian expert
510 Points
```										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...

```
6 years ago
Akarsh Gaurav
24 Points
```										y=f(x)=cosx.cos2x.cos4x.cos8x.cos16x         =2sinx.cosx.cos2x.cos4x.cos8x.cos16x/2sinx                      [${\color{Red} \because }\displaystyle$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${\color{Red}\therefore }\displaystyle$y=f(x)=sin32x/32sinxdifferentiate both sides w.r.to x,dy/dx=d(sin32x/32sinx)/dxdifferentiate it you will getthe answer. THANK YOU
```
6 months ago
Nitish jha
28 Points
```										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
```
2 months ago
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