If y = xlog xlog log x,then dydx=?
xlogxlog logx1xlogxlogx+log logx+log logx1x+1xlogx
xlogxlog logxlog (log x)2logx+1x
xlogxxlogxlog logxx1logx+1
ylogyxlogx2loglogx+1
Explanation for the correct option:
Step 1: Simplifying the given equation:
Given that,
y = xlogxlog log x
By taking log on both sides,
logy=logxlogxloglogx =logxloglogx.logx [∵ logab=bloga]
Now, by taking log again on both sides,
loglogy=loglogxloglogx.logx =loglogxloglogx+loglogx [∵ log(ab)=loga+logb] =loglogx.loglogx+loglogx=loglogx2+loglogx ...1
Step 2: Find the dydx:
Differentiate equation 1 both sides with respect to x, we get
1logy.1ydydx=2loglogx.1logx.1x+1logx×1x [∵ ddx(logx)=1x]⇒ 1ylogy.dydx=1xlogx2loglogx+1⇒ dydx=ylogyxlogx2loglogx+1
Hence, the correct option is D.
Complete the Tables
Expression
Term with factor of x
Coefficient of x
3x-5
3x
3
8-x+y
2z-5xz