Question

find the differential cofficient of log(log x) ?​

Answer 1

Answer:

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))

Answer 2

Answer:

1(xlogx)

Step-by-step explanation:

y=log(logx)

let u =log x

du/DX=1/x

thus,y=log u

defferentiate the above equation wet x

dy(DX=1/u*du/DX

=1/logx*1/x

=1(xlogx)