Differentiate cosx/logx, x>0

Let , y= cosx/logx


If y= u/v , dy/dx =( u`v- uv`)/v^2


dy/dx = (d/dxcosx)logx - cosx d/dx(logx)


= (-sinxlogx - cosx * 1/x )/( logx)^2


= xsinxlogx + cosx/ x (logx)^2 Ans.

the derivative of u/v is (u'v-uv')/v^2

u=cosx
u'=-sinx

v=logx
10^v=x
vln10=lnx
ln10v'=1/x
v'=1/(xln10)

so:

[-sinxlogx-cosx/(xln10)^2]/(logx)^2

[(Sec x)^2log x - 1/x cos x ]/ (log x )^2

(sinx*logx-cosx/x)/(logx)^2