Question

How do you find the derivative of `g(t) = (ln(kt)+2t) / (ln(kt)-2t)`?

Answer 1

Here, we'll use the quotient rule, which states that, be `y=f(x)/g(x)`,

`(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/g(x)^2`

Let's just find the derivatives and use the function properly:

• `f(x)=lnkt+2t`

• `f'(x)` will demand chain rule, which states: `(dy)/(dx)=(dy)/(du)(du)/(dx)`

Thus, renaming `u=kt`, we have `f'(x)=(1/u)(k)+2=(cancel(k)/(cancel(k)t))+2=1/t+2`

• `g(x)=lnkt-2t`

• `g'(x)` follows the logic for `f'(x)`: `g'(x)=1/t-2`

Now, applying the quotient rule:

`(dy)/(dx)=((1/t+2)(lnkt-2t)-(lnkt+2t)(1/t-2))/(lnkt-2t)^2`

`(dy)/(dx)=(cancel((lnkt)/t)-2+2lnktcancel(-4)tcancel(-(lnkt)/t)+2lnkt-2cancel(+4t))/(lnkt-2t)^2`

`(dy)/(dx)=(4lnkt-4)/(lnkt-2t)^2=(4(lnkt-1))/(lnkt-2t)^2`