Question

Hi!❤️

y = log(logx)

Differentiate with respect to x​

Answer 1

Answer:

hlwww dear.........

Step-by-step explanation:

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

Differentitate the above equation wrt x,

dy/dx = 1/u *du/dx

= 1/logx *1/x

= 1/(xlogx)

Hope it's help you

Thank you...............

Answer 2

Heya Buddhu :)♥

Answer:-

Y =log (logx )

differentiate wrt x

dy/dx ={1/logx}1/x =(xlogx)^-1

again differentiate wrt x

d^2y/dx^2 =-(xlogx)^-2 {logx + x 1/x }

= -(1 + logx )/(x. logx)^2