A curve is represented parametrically by the equations `x=f(t)=a^(In(b^t))and y=g(t)=b^(-In(a^(t)))a,bgt0 and a ne 1, b ne 1" Where "t in R.`
The value of `(d^(2)y)/(dx^(2))` at the point where f(t)=g(t) is

Options

A. `(1)/(f(t)^(2))`
B. `-(g(t))^(2)`
C. `(-g(t))/f(t)`
D. `(-f(t))/(g(t))`

Explanation

`x=f(t)=a^(In(b))=a^(t" In "b)" (1)"`
`y=g(t)=b^(-In(a^(t)))=(b^(In a))^(-t)=(a^(In b))^(-t)=a^(-t" In b")`
`therefore" "y=g(t)=a^(In(b^(-t)))=f(-t)" (2)"`
From equations (1), and (2),
`xy=1`
`y=(1)/(x)`
`therefore" "(dy)/(dx)=-(1)/(x^(2))=-(1)/(f^(2)(t))`
`"Also, "xy=1 rArr-(1)/(f^(2)(t))=-g^(2)(t)`
`therefore" "(dy)/(dx)=-(1)/(x^(2))=-(y^(2))/(1)`
`"Also, "xy=1 rArr(dy)/(dx)=-(y)/(x)=-(g(t))/(f(t))`