Find the derivative of `y = ln 2x`

Answer

`ator; b to p ; c to s ; d toq`

Explanation

a. if `veca,vecb and vecc` are mutaully perpendicular , then
`[vecaxx vecb , vecbxx vecc,veccxxveca]=[veca vecb vecc]^(2)`
`(|veca||vecb||vecc|)^(2)=16`
b. Given `veca and vecb` are two unit vectors, i.e., `|veca|=|vecb|=1` and angle between then is `pi//3`
`sin theta= (|vecaxxvecb|)/(|veca ||vecb|) Rightarrow sin"" pi/3=|vecaxxvecb|`
`sqrt3/2 |veca xx vecb|`
Now
`[veca vecb+vecaxxvecb vecb] = [veca vecb vecb] + [veca veca xxvecb vecb]`
` = 0+ [veca veca xx vecb vecb]`
`= (veca xx vecb). (vecb xx veca)`
`- (veca xx vecb). (veca xx vecb)`
`-|veca xx vecb|^(2)`
`-3/4`
c. if `vecb and vecc` are othogonal `vecb. vecc =0`
Also, if is given that `vecb xx vecc=veca` Now,
`[veca+vecb +vecc veca +vecb vecb + vecc]`
`[veca veca+vecb vecb +vecc] + [vecb +vecc vec a +vecb vecb +vecc]`
`[veca vecb vecc]`
` = veca.(vecb xx vecc)`
`veca. veca = |veca|^(2)=1` ( because `veca` is a unit vector)
d. ` [vecx vecy veca ] =0`
Therefore, `vecx, vecy and veca` are coplanar,
`[vecx vecy vecb] =0`
Therefore, `vecx, vecy and vecb` are coplanar,
Also , `[veca vecb vecc] =0`
Therefore, `veca, vecb and vecc` are coplanar,
From (i), (ii) and (iii) ,
` vecx ,vecy and vecc` are coplanar, therefore,
`[vecx vecy vecc] =0`