differentiate the following log(logx)
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The function we have is log(logx)
y = log(logx)
We make use of chain rule
Take u= logx
du / dx =1/x
Thus y= logu
Differentiate the above equation wrt x,
dy/dx = 1/u * du/dx
= 1/logx * 1/x = 1/ (xlogx
y = log(logx)
We make use of chain rule
Take u= logx
du / dx =1/x
Thus y= logu
Differentiate the above equation wrt x,
dy/dx = 1/u * du/dx
= 1/logx * 1/x = 1/ (xlogx
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